Two-Dimensional Motion

1. Vectors in Motion

Scalars and vectors

Physical quantities can be classified as scalars or vectors.

A scalar quantity has magnitude only.

A vector quantity has both magnitude and direction.

Magnitude describes the size or amount of a quantity. For example, a velocity might have a magnitude of 12 m/s.

A scalar can usually be described with a number and a unit:

Speed = 12 m/s

A vector also requires direction:

Velocity = 12 m/s east

Direction is essential. Two objects travelling at 12 m/s in opposite directions have equal speeds but different velocities.

Distance and displacement

Distance is the total length of the path travelled. It is a scalar.

Displacement is the change in position from the starting point to the finishing point. It is a vector.

Imagine a student walking:

• 5 m east.
• Then 2 m west.

Total distance:

d = 5 + 2

d = 7 m

Taking east as positive, displacement is:

s = +5 − 2

s = +3 m

The student’s displacement is 3 m east.

Distance depends on the complete path. Displacement depends only on the initial and final positions.

A winding route may have a large distance but a much smaller displacement between its starting and finishing points.

Speed and velocity

Speed describes how quickly distance is travelled:

Average speed = total distance/total time

Velocity describes how quickly displacement changes:

Average velocity = displacement/time

Speed is scalar. Velocity is vector.

Worked example

A runner completes one 400 m lap in 80 seconds and finishes at the starting line.

Average speed:

Average speed = 400/80

Average speed = 5 m/s

The runner’s displacement is zero, so:

Average velocity = 0/80

Average velocity = 0 m/s

The runner was moving throughout the lap, but the average velocity is zero because the final position equals the initial position.

Representing a vector

A vector can be represented using an arrow.

The arrow shows:

• Length: The vector’s magnitude according to a chosen scale.
• Arrowhead: The vector’s direction.
• Orientation: The line along which the vector acts.

For example, a scale diagram might use:

1 cm represents 5 m/s

A velocity of 15 m/s east would be drawn as a 3 cm arrow pointing east.

A velocity of 10 m/s north would be drawn as a 2 cm arrow pointing north.

The scale must be stated so that the vector’s magnitude can be recovered from the diagram.

Describing vector directions

Vectors can be described using compass directions:

• North.
• South.
• East.
• West.
• Northeast and other combinations.

They can also be described using angles:

• 30° north of east.
• 20° west of north.
• A bearing of 065°.
• 40° above the horizontal.

30° north of east means begin facing east and rotate 30° towards north.

30° east of north means begin facing north and rotate 30° towards east.

These descriptions are not equivalent.

Vector notation

Vectors may be written using:

• A bold symbol, such as v.
• A symbol with an arrow above it.
• Component notation, such as ⟨vₓ, vᵧ⟩.

The magnitude of a vector may be written as:

|v|

For example, if:

v = ⟨6, 8⟩ m/s

then its magnitude is:

|v| = √(6² + 8²)

|v| = √100

|v| = 10 m/s

The vector and its magnitude are different:

• v includes direction.
• |v| is a non-negative scalar.

Adding vectors

When several vectors act in sequence, they can be combined to find a resultant vector.

The resultant has the same overall effect as the original vectors together.

For displacement:

Resultant displacement = displacement₁ + displacement₂ + …

For velocity:

Resultant velocity = velocity₁ + velocity₂ + …

Vector addition must account for direction.

Head-to-tail addition

To add vectors graphically:

1. Choose and state a scale.
2. Draw the first vector accurately.
3. Place the tail of the second vector at the head of the first.
4. Continue this process for any additional vectors.
5. Draw the resultant from the tail of the first vector to the head of the final vector.
6. Measure the resultant’s length and direction.

The vectors can be added in either order without changing the resultant:

A + B = B + A

The first diagram shows a 3 m eastward displacement followed by a 4 m northward displacement. The resultant is the diagonal vector from the starting point to the finishing point.

Worked example: perpendicular displacements

A person walks 3 m east and then 4 m north.

These displacements form a right-angled triangle.

Use the Pythagorean theorem to find the resultant magnitude:

R² = 3² + 4²

R² = 9 + 16

R = 5 m

Find the angle measured north of east:

tan θ = opposite/adjacent

tan θ = 4/3

θ = tan⁻¹(4/3)

θ ≈ 53°

The resultant displacement is:

5 m at 53° north of east

The total distance travelled is 7 m, while the displacement magnitude is 5 m.

Adding vectors in the same dimension

Vectors acting along one straight line can be added using positive and negative signs.

Choose right as positive.

Worked example

A trolley moves:

• 12 m right.
• Then 5 m left.
• Then 3 m right.

Write the displacements with signs:

s₁ = +12 m
s₂ = −5 m
s₃ = +3 m

Add them:

s = 12 − 5 + 3

s = +10 m

The resultant displacement is 10 m to the right.

The total distance is:

d = 12 + 5 + 3

d = 20 m

Adding non-perpendicular vectors graphically

When vectors are not perpendicular, they can still be added using a scale diagram.

For example, to add:

• 8 m east.
• 6 m at 40° north of east.

Draw the first vector, then place the second vector head-to-tail at the correct angle. Draw and measure the resultant.

The accuracy depends on:

• The size of the scale drawing.
• Accurate angle measurement.
• Thin, precise lines.
• Careful measurement of the resultant.

A large scale usually produces a more accurate answer.

For exact results, vectors can be resolved into components and added algebraically.

Subtracting vectors

Vector subtraction can be rewritten as addition of the opposite vector:

A − B = A + (−B)

The vector −B has the same magnitude as B but points in the opposite direction.

To subtract graphically:

1. Draw vector A.
2. Reverse vector B to obtain −B.
3. Add A and −B using the head-to-tail method.

If vectors A and B are drawn from the same origin, A − B is the vector from the head of B to the head of A.

Change in velocity

Acceleration depends on the change in velocity:

a = Δv/Δt

The change in velocity is:

Δv = v − u

Because velocity is a vector, this subtraction must account for direction.

Worked example: straight-line velocity change

A car’s velocity changes from +5 m/s to +17 m/s in 4 seconds.

Δv = v − u

Δv = 17 − 5

Δv = +12 m/s

Acceleration:

a = 12/4

a = +3 m/s²

Worked example: reversing direction

A ball’s velocity changes from +6 m/s to −4 m/s in 2 seconds.

Δv = v − u

Δv = −4 − (+6)

Δv = −10 m/s

Acceleration:

a = −10/2

a = −5 m/s²

The velocity did not change by only 2 m/s. The object passed through zero velocity and reversed direction.

Direction changes create acceleration

An object can accelerate even when its speed remains constant.

Suppose a car travels around a circular track at a constant speed. Its direction changes continuously, so its velocity changes continuously.

Therefore, the car accelerates even though its speedometer reading may remain constant.

In circular motion, instantaneous velocity is tangent to the path, while acceleration points towards the centre of the circle.

Resolving vectors into components

To resolve a vector means to separate it into perpendicular components.

In two-dimensional motion, these are usually:

• A horizontal component.
• A vertical component.

The components combine to produce the original vector.

For a vector V making an angle θ above the positive horizontal:

Vₓ = V cos θ

Vᵧ = V sin θ

These equations apply when θ is measured from the horizontal.

If the angle is measured from the vertical, the sine and cosine relationships exchange roles. A diagram helps prevent mistakes.

Worked example: resolving velocity

An object moves at 10 m/s at 35° above the horizontal.

Horizontal component:

vₓ = v cos θ

vₓ = 10 cos 35°

vₓ ≈ 8.19 m/s

Vertical component:

vᵧ = v sin θ

vᵧ = 10 sin 35°

vᵧ ≈ 5.74 m/s

The velocity can be written in component form:

v = ⟨8.19, 5.74⟩ m/s

These components act together. They do not represent two separate velocities occurring at different times.

Reconstructing a vector from components

If the components are known, use the Pythagorean theorem to find the magnitude:

V = √(Vₓ² + Vᵧ²)

Find its direction using:

tan θ = Vᵧ/Vₓ

Therefore:

θ = tan⁻¹(Vᵧ/Vₓ)

Worked example

A drone’s velocity components are:

vₓ = 12 m/s east
vᵧ = 5 m/s north

Magnitude:

v = √(12² + 5²)

v = √169

v = 13 m/s

Direction:

θ = tan⁻¹(5/12)

θ ≈ 22.6°

The drone’s velocity is:

13 m/s at 22.6° north of east

Always check the signs of the components to determine the correct quadrant.

Signs and quadrants

If east and north are positive:

A calculator’s inverse tangent may return an angle that does not identify the correct quadrant. Inspect the component signs and describe the direction clearly.

For example:

vₓ = −6 m/s
vᵧ = +8 m/s

The vector points northwest.

Its magnitude is:

v = √[(−6)² + 8²]

v = 10 m/s

The reference angle is:

θ = tan⁻¹(8/6)

θ ≈ 53°

A clear direction is:

53° north of west

Adding vectors using components

Component addition is often more accurate than a scale drawing.

To add several vectors:

1. Resolve each vector into horizontal and vertical components.
2. Add all horizontal components.
3. Add all vertical components.
4. Reconstruct the resultant magnitude and direction.

Rₓ = Aₓ + Bₓ + …

Rᵧ = Aᵧ + Bᵧ + …

Then:

R = √(Rₓ² + Rᵧ²)

Worked example: adding angled displacements

A hiker walks:

• 6.0 km east.
• Then 4.0 km at 30° north of east.

Resolve the second displacement:

Bₓ = 4.0 cos 30°

Bₓ ≈ 3.46 km

Bᵧ = 4.0 sin 30°

Bᵧ = 2.00 km

Add components:

Rₓ = 6.0 + 3.46

Rₓ = 9.46 km

Rᵧ = 0 + 2.00

Rᵧ = 2.00 km

Magnitude:

R = √(9.46² + 2.00²)

R ≈ 9.67 km

Direction:

θ = tan⁻¹(2.00/9.46)

θ ≈ 11.9°

The resultant displacement is approximately:

9.67 km at 11.9° north of east

Vectors in projectile motion

Projectile motion can be analyzed by separating the velocity into horizontal and vertical components.

If air resistance is ignored:

• Horizontal acceleration is zero.
• Horizontal velocity remains constant.
• Vertical acceleration is caused by gravity.
• Horizontal and vertical motion occur during the same time interval.

The horizontal and vertical components change differently during flight, but together they produce the projectile’s curved path.

Worked example: initial projectile components

A ball is kicked at 20 m/s at 40° above the horizontal.

Horizontal component:

uₓ = 20 cos 40°

uₓ ≈ 15.3 m/s

Vertical component:

uᵧ = 20 sin 40°

uᵧ ≈ 12.9 m/s

Ignoring air resistance:

• The horizontal velocity remains approximately 15.3 m/s.
• The vertical velocity decreases as gravity acts downwards.
• At the highest point, vertical velocity is zero.
• The ball still has horizontal velocity at the highest point.

The total velocity is not zero at the top of the trajectory.

Relative velocity

Relative velocity describes the velocity of one object as observed from another moving object.

For objects A and B:

vₐ relative to B = vₐ − vᵦ

Worked example: vehicles moving in the same direction

Car A travels east at 25 m/s. Car B travels east at 18 m/s.

Velocity of A relative to B:

vₐᵦ = 25 − 18

vₐᵦ = 7 m/s east

A passenger in B sees A move forwards at 7 m/s.

Worked example: vehicles moving in opposite directions

Car A travels east at +25 m/s. Car B travels west at −18 m/s.

Velocity of A relative to B:

vₐᵦ = 25 − (−18)

vₐᵦ = 43 m/s east

Their separation changes at 43 m/s.

Relative velocity with wind and water

A boat’s velocity relative to the water combines with the water’s velocity relative to the ground.

Boat velocity relative to ground = boat velocity relative to water + water velocity relative to ground

The same principle applies to aircraft and wind.

Pilots and boat operators must account for the motion of the air or water to reach the intended destination.

Worked example: boat crossing a river

A boat moves north through the water at 4 m/s. The river current flows east at 3 m/s.

The velocity components relative to the ground are:

vₓ = 3 m/s east
vᵧ = 4 m/s north

Magnitude:

v = √(3² + 4²)

v = 5 m/s

Direction:

θ = tan⁻¹(3/4)

θ ≈ 36.9° east of north

The boat’s ground velocity is 5 m/s at 36.9° east of north.

If the river is 80 m wide, the crossing time depends on the northward component:

t = 80/4

t = 20 s

During that time, the current carries the boat east:

sₓ = 3(20)

sₓ = 60 m

The total speed of 5 m/s should not be used to calculate the crossing time because only the northward component carries the boat across the river.

Acceleration as a vector

Acceleration describes the change in the velocity vector:

a = Δv/Δt

Because velocity has magnitude and direction, acceleration may result from:

• A change in speed.
• A change in direction.
• Changes in both speed and direction.

Acceleration can also be resolved into components:

a = ⟨aₓ, aᵧ⟩

For projectile motion without air resistance:

aₓ = 0

aᵧ = −g, if upward is positive

For circular motion, acceleration points towards the centre of the path.

Multiplying vectors by time

For constant velocity:

s = vt

Multiplying a velocity vector by a positive time interval produces a displacement vector in the same direction.

For example:

v = ⟨4, 3⟩ m/s
t = 5 s

Then:

s = ⟨4, 3⟩(5)

s = ⟨20, 15⟩ m

The displacement magnitude is:

|s| = √(20² + 15²)

|s| = 25 m

Its direction is:

θ = tan⁻¹(15/20)

θ ≈ 36.9° north of east

Vector diagrams and scale

A high-quality vector diagram should include:

• A stated scale.
• Clearly labelled vectors.
• Arrowheads showing direction.
• Accurate angles.
• Suitable units.
• The resultant beginning and ending at the correct points.

When solving graphically, do not calculate the resultant by adding the arrow lengths unless the vectors point along the same line in the same direction.

A graphical answer is usually approximate. Component calculations can produce a more precise result.

Common misconceptions

• “A vector is any quantity with a positive or negative sign.” A vector has both magnitude and direction.
• “Distance and displacement are always equal.” They are equal only for suitable one-direction paths.
• “Speed and velocity are interchangeable.” Velocity includes direction.
• “Vectors are added by adding their magnitudes.” Direction must be considered.
• “The resultant is drawn by joining the two arrowheads.” For head-to-tail addition, it runs from the first tail to the final head.
• “Sine always gives the vertical component.” This depends on where the angle is measured.
• “At the top of a projectile’s path, velocity is zero.” Only its vertical velocity is zero if horizontal motion continues.
• “Constant speed means zero acceleration.” A changing direction produces acceleration.
• “A negative vector has a negative magnitude.” Magnitude is non-negative; the sign indicates direction along an axis.

Did you know?

Air-traffic controllers use vector calculations to separate an aircraft’s motion through the air from the motion of the surrounding air mass.

An aircraft may point in one direction while travelling over the ground in another direction because crosswind changes its resultant velocity.

Key terms

• Scalar: A quantity with magnitude only.
• Vector: A quantity with magnitude and direction.
• Magnitude: The size of a scalar or vector quantity.
• Direction: The orientation in which a vector acts.
• Resultant vector: A single vector with the same effect as two or more combined vectors.
• Component: Part of a vector acting along a selected axis.
• Head-to-tail method: A graphical method of adding vectors sequentially.
• Resolution: The process of separating a vector into perpendicular components.
• Relative velocity: The velocity of one object measured from the viewpoint of another.
• Displacement: Change in position, including direction.
• Projectile: An object moving under gravity after being launched.
• Trajectory: The path followed by a moving object.

Key takeaways

• Scalars have magnitude; vectors have magnitude and direction.
• Distance and speed are scalars, while displacement, velocity and acceleration are vectors.
• Represent a vector with a scaled arrow.
• Add vectors using the head-to-tail method or components.
• Subtract a vector by adding its opposite.
• Use cosine and sine to find components when the angle is measured from the horizontal.
• Add horizontal components separately from vertical components.
• Reconstruct magnitude using the Pythagorean theorem.
• Changes in speed or direction both produce acceleration.
• Vector methods are essential for projectiles, navigation and relative motion.

2. Projectile Motion

What is projectile motion?

A projectile is an object that moves through the air after being launched and is then influenced mainly by gravity.

Examples include:

• A ball that has been thrown or kicked.
• A basketball travelling towards the hoop.
• A stone launched from a cliff.
• Water leaving a fountain.
• A package released from a moving aircraft.

Projectile motion is two-dimensional because the object moves both horizontally and vertically.

When air resistance is ignored:

• Gravity is the only force acting after launch.
• Horizontal acceleration is zero.
• Vertical acceleration is constant and directed downwards.
• The resulting trajectory is a parabola.

Although the path is curved, it can be understood by separating the motion into horizontal and vertical components.

Assumptions of the basic projectile model

Introductory projectile problems usually assume:

• Air resistance is negligible.
• Gravitational acceleration is constant.
• Earth’s surface is locally flat.
• The object does not produce thrust after launch.
• The object can be represented as a single point.
• Earth’s rotation is ignored.

Near Earth’s surface, gravitational acceleration is approximately:

g = 9.8 m/s² downward

Some problems use:

g = 10 m/s²

Always use the value provided in the question.

Horizontal and vertical motion

Projectile motion can be separated into two perpendicular parts.

Horizontal motion

With no air resistance, there is no horizontal force and therefore no horizontal acceleration:

aₓ = 0

The horizontal velocity remains constant:

vₓ = uₓ

Horizontal displacement is:

x = uₓt

Vertical motion

Gravity produces constant vertical acceleration.

If upwards is positive:

aᵧ = −g

The vertical velocity changes according to:

vᵧ = uᵧ − gt

The vertical displacement is:

y = uᵧt − ½gt²

Another useful equation is:

vᵧ² = uᵧ² − 2gy

Only vertical quantities should be used in the vertical equations.

The motions occur at the same time

The horizontal and vertical motions can be calculated separately, but they happen during the same time interval.

Time connects the components:

• The time taken to move horizontally is the same as the time spent rising or falling.
• Horizontal motion does not create vertical acceleration.
• Vertical motion does not change horizontal velocity in the ideal model.
• The two components combine to produce the actual trajectory.

Gravity does not “wait” until the projectile stops moving horizontally. It acts vertically throughout the flight.

A projectile’s changing velocity

The diagram represents an object launched at 20 m/s at 45°.

• The horizontal component remains constant.
• The vertical component decreases steadily because of gravity.
• At the highest point, vertical velocity is zero.
• After the highest point, vertical velocity is negative.
• The total velocity is tangent to the curved trajectory.

Resolving the initial velocity

If a projectile is launched at speed u and angle θ above the horizontal, resolve its initial velocity into components:

uₓ = u cos θ

uᵧ = u sin θ

These relationships apply when θ is measured from the horizontal.

Worked example

A ball is launched at 20 m/s at 45° above the horizontal.

Horizontal component:

uₓ = 20 cos 45°

uₓ ≈ 14.1 m/s

Vertical component:

uᵧ = 20 sin 45°

uᵧ ≈ 14.1 m/s

The initial velocity can be written in component form:

u = ⟨14.1, 14.1⟩ m/s

The horizontal and vertical components are equal because the launch angle is 45°.

Horizontal launch

A horizontally launched projectile has:

uₓ = launch speed

uᵧ = 0

It begins falling immediately, even though its initial vertical velocity is zero.

The horizontal gaps between successive positions are equal in the ideal model because horizontal velocity is constant. The vertical gaps grow because gravity increases the downward velocity.

Worked example: ball rolling from a table

A ball rolls horizontally from a 1.25 m high table at 4.0 m/s. Calculate the time it takes to reach the floor and its horizontal range. Use g = 10 m/s².

Choose downward as positive for the vertical calculation.

Vertical information:

uᵧ = 0 m/s
aᵧ = 10 m/s²
y = 1.25 m
t = ?

Use:

y = uᵧt + ½aᵧt²

Substitute:

1.25 = 0 + ½(10)t²

1.25 = 5t²

t² = 0.25

t = 0.50 s

Now analyze the horizontal motion:

uₓ = 4.0 m/s
t = 0.50 s

x = uₓt

x = 4.0(0.50)

x = 2.0 m

The ball lands 2.0 m horizontally from the table.

Why horizontal speed does not affect fall time

If two objects leave the same height at the same time with equal initial vertical velocities, they land at the same time when air resistance is ignored.

One object could be dropped vertically while the other is launched horizontally. Their horizontal motions differ, but their vertical motions are identical.

The horizontally launched object travels farther sideways when its horizontal speed is greater, but it does not take longer to fall.

This is an example of the independence of perpendicular motion components.

Finding the trajectory equation

The time equations can be combined to show why the trajectory is parabolic.

Horizontal motion:

x = uₓt

Therefore:

t = x/uₓ

Vertical motion:

y = uᵧt − ½gt²

Substitute t = x/uₓ:

y = (uᵧ/uₓ)x − [g/(2uₓ²)]x²

For launch speed u at angle θ:

uᵧ/uₓ = tan θ

uₓ = u cos θ

Therefore:

y = x tan θ − [gx²/(2u²cos²θ)]

This equation has the form:

y = Ax − Bx²

It is a quadratic equation, so its graph is a parabola.

Projectile velocity at any time

The horizontal component remains constant:

vₓ = uₓ

The vertical component changes:

vᵧ = uᵧ − gt

The resultant speed is:

v = √(vₓ² + vᵧ²)

The direction relative to the horizontal can be found using:

tan θ = vᵧ/vₓ

The sign of vᵧ indicates whether the projectile is rising or falling.

Worked example: velocity during flight

A ball is launched at 25 m/s at 30° above the horizontal. Find its velocity components after 1.0 s. Use g = 9.8 m/s².

Initial horizontal component:

uₓ = 25 cos 30°

uₓ ≈ 21.7 m/s

Initial vertical component:

uᵧ = 25 sin 30°

uᵧ = 12.5 m/s

After 1.0 s:

vₓ = 21.7 m/s

vᵧ = uᵧ − gt

vᵧ = 12.5 − 9.8(1.0)

vᵧ = 2.7 m/s

Resultant speed:

v = √(21.7² + 2.7²)

v ≈ 21.9 m/s

Direction:

θ = tan⁻¹(2.7/21.7)

θ ≈ 7.1° above the horizontal

The ball is still rising because its vertical velocity is positive.

Maximum height

At the highest point:

vᵧ = 0

The horizontal velocity usually remains non-zero.

To find the time to maximum height:

0 = uᵧ − gt

Therefore:

t_up = uᵧ/g

To find the maximum height above the launch point:

vᵧ² = uᵧ² − 2gH

At the top, vᵧ = 0:

H = uᵧ²/(2g)

Worked example: maximum height

A ball is launched at 20 m/s at 45°. Use g = 9.8 m/s².

The initial vertical velocity is:

uᵧ = 20 sin 45°

uᵧ ≈ 14.1 m/s

Time to maximum height:

t_up = 14.1/9.8

t_up ≈ 1.44 s

Maximum height:

H = 14.1²/[2(9.8)]

H ≈ 10.2 m

The ball reaches approximately 10.2 m above its launch point.

Total flight time

If the projectile lands at the same height from which it was launched, the downward motion is symmetrical with the upward motion in the ideal model.

Therefore:

Total flight time = 2uᵧ/g

For the ball launched at 20 m/s at 45°:

T = 2(14.1)/9.8

T ≈ 2.89 s

This shortcut applies only when the landing height equals the launch height.

If the projectile lands at a different height, solve the vertical displacement equation instead.

Horizontal range

The range is the horizontal displacement from launch to landing.

For a projectile that lands at its launch height:

R = uₓT

Substituting the component and flight-time expressions gives:

R = u²sin(2θ)/g

For the 20 m/s projectile at 45°:

R = 20²sin 90°/9.8

R = 400/9.8

R ≈ 40.8 m

This agrees with the trajectory diagram.

The effect of launch angle

For a fixed launch speed and equal launch and landing heights:

• A small angle produces a low, flatter trajectory.
• A large angle produces a high, steep trajectory.
• The maximum ideal range occurs at 45°.
• Complementary angles have the same range.

For example:

• 30° and 60° produce the same ideal range.
• The 30° projectile travels lower and faster horizontally.
• The 60° projectile rises higher and remains airborne longer.

These conclusions assume no air resistance and equal launch and landing heights. Real sports projectiles may have a different best angle.

Worked example: angled launch

A football is kicked from ground level at 18 m/s at 35° above the horizontal. Calculate its flight time, maximum height and range. Use g = 9.8 m/s² and ignore air resistance.

Resolve the initial velocity

uₓ = 18 cos 35°

uₓ ≈ 14.7 m/s

uᵧ = 18 sin 35°

uᵧ ≈ 10.3 m/s

Calculate flight time

Because the ball lands at its launch height:

T = 2uᵧ/g

T = 2(10.3)/9.8

T ≈ 2.11 s

Calculate maximum height

H = uᵧ²/(2g)

H = 10.3²/[2(9.8)]

H ≈ 5.41 m

Calculate range

R = uₓT

R = 14.7(2.11)

R ≈ 31.0 m

The football remains airborne for approximately 2.11 s, rises 5.41 m and travels 31.0 m horizontally.

Launching from an elevated position

When launch and landing heights are different, the upward and downward sections are not symmetrical.

Use the vertical displacement equation:

y = uᵧt − ½gt²

The equation may produce two mathematical solutions. Select the time that matches the physical situation.

Worked example: launch from a cliff

A stone is launched horizontally at 12 m/s from a cliff 45 m above the water. Use g = 10 m/s².

Choose the launch point as y = 0 and downward as positive.

Vertical motion:

uᵧ = 0 m/s
aᵧ = 10 m/s²
y = 45 m

Use:

y = uᵧt + ½aᵧt²

45 = 0 + 5t²

t² = 9

t = 3 s

Horizontal range:

x = uₓt

x = 12(3)

x = 36 m

Vertical velocity at impact:

vᵧ = uᵧ + aᵧt

vᵧ = 0 + 10(3)

vᵧ = 30 m/s downward

Horizontal velocity at impact:

vₓ = 12 m/s

Resultant impact speed:

v = √(12² + 30²)

v ≈ 32.3 m/s

Predicting the shape of a trajectory

To predict a trajectory, consider the initial components and acceleration.

Larger horizontal velocity

A larger uₓ causes the projectile to travel farther horizontally during the same flight time.

Larger vertical velocity

A larger uᵧ increases:

• Time spent rising.
• Maximum height.
• Total flight time, when landing height is unchanged.

Greater gravitational acceleration

A greater g causes the projectile to:

• Lose upward velocity more quickly.
• Reach a lower maximum height.
• Spend less time in the air.
• Travel a shorter horizontal range for the same launch velocity.

Launching from a greater height

A greater launch height normally increases the flight time and horizontal range.

Real-world applications

Sports

Projectile analysis helps athletes and coaches study:

• Basketball shots.
• Football kicks.
• Javelin throws.
• Long jumps.
• Golf shots.
• Volleyball serves.

Real sports motion is more complicated because air resistance, spin, object shape and release height affect the trajectory.

Engineering

Engineers apply projectile models to:

• Water fountains.
• Material released from conveyor belts.
• Fire-hose streams.
• Vehicle safety testing.
• Robotics.
• Aerospace systems.

Forensic and safety analysis

Trajectory calculations can help investigators estimate:

• The launch position.
• Initial speed.
• Direction of motion.
• Time in flight.
• Possible landing region.

Such calculations require reliable measurements and an appropriate model.

Air resistance

Air resistance changes both horizontal and vertical motion.

With air resistance:

• Horizontal velocity decreases.
• The path is no longer a perfect parabola.
• The projectile normally travels a shorter range.
• Ascent and descent are not perfectly symmetrical.
• Impact speed may be lower than the ideal prediction.

The effect is especially significant for objects that are light, large, irregular or moving quickly.

Examples include:

• Feathers.
• Shuttlecocks.
• Footballs.
• Paper projectiles.
• Parachutists.

The ideal model works best when drag is small compared with the object’s weight.

Graphs of projectile motion

Horizontal displacement–time graph

Horizontal displacement increases linearly:

x = uₓt

Its constant gradient is uₓ.

Vertical displacement–time graph

Vertical displacement follows a quadratic relationship:

y = uᵧt − ½gt²

Its graph is a parabola.

Horizontal velocity–time graph

Horizontal velocity is constant, so the graph is horizontal.

Vertical velocity–time graph

Vertical velocity decreases linearly:

vᵧ = uᵧ − gt

Its gradient is −g.

Vertical acceleration–time graph

Vertical acceleration remains at −g, so the graph is a horizontal line below the time axis when upwards is positive.

Solving projectile problems

A reliable method is:

1. Draw the trajectory and coordinate axes.
2. Choose positive horizontal and vertical directions.
3. Resolve the initial velocity into components.
4. List the horizontal variables.
5. List the vertical variables.
6. Use vertical motion to find time where necessary.
7. Use the shared time in the horizontal equation.
8. Recombine velocity components if a resultant velocity is required.
9. Interpret the answer using units and direction.
10. Check whether the assumptions are reasonable.

Keeping horizontal and vertical calculations in separate columns can prevent variables from being mixed.

Common misconceptions

• “Gravity reduces horizontal velocity.” In the ideal model, gravity acts vertically.
• “The projectile has no acceleration at its highest point.” Its acceleration remains g downward.
• “The projectile stops at the highest point.” Only its vertical velocity is zero; horizontal velocity continues.
• “Horizontal and vertical motion happen one after the other.” They occur simultaneously.
• “A horizontal launch has no vertical motion at first.” It begins falling immediately.
• “The launch speed can be used directly in both directions.” Resolve it into components first.
• “The flight is always symmetrical.” Symmetry requires equal launch and landing heights and negligible air resistance.
• “The best range angle is always 45°.” This is true only for the ideal equal-height model.

Did you know?

Astronauts in orbit are continually falling towards Earth. Their large horizontal velocity causes Earth’s curved surface to fall away beneath them at the same rate.

An orbit can therefore be understood as an extended form of projectile motion motion, although accurate orbital calculations require gravitational models beyond constant g.

Key terms

• Projectile: An object moving under gravity after launch.
• Projectile motion: Two-dimensional motion combining horizontal and vertical components.
• Trajectory: The path followed by a projectile.
• Component: Part of a vector acting along a chosen axis.
• Horizontal range: Horizontal displacement from launch to landing.
• Maximum height: Greatest vertical position above a reference level.
• Flight time: Total time between launch and landing.
• Launch angle: Direction of the initial velocity relative to the horizontal.
• Free fall: Motion in which gravity is the only significant force.
• Parabola: The ideal curved trajectory of a projectile.
• Air resistance: A drag force opposing motion through air.
• Resultant velocity: The vector combination of horizontal and vertical velocities.

Key takeaways

• Projectile motion combines horizontal and vertical motion.
• The two components can be analyzed independently because gravity acts vertically.
• Horizontal velocity remains constant when air resistance is ignored.
• Vertical acceleration remains equal to g downward.
• Resolve an angled launch using uₓ = u cos θ and uᵧ = u sin θ.
• At maximum height, vᵧ = 0 but acceleration is still downward.
• Time connects the horizontal and vertical calculations.
• The ideal projectile trajectory is parabolic.
• Launch speed, angle, height and gravity determine the trajectory.
• Air resistance makes real trajectories differ from the ideal model.

3. Free Fall

What is free fall?

An object is in free fall when gravity is the only force acting on it.

A freely falling object may be:

• Moving downwards after being dropped.
• Moving downwards after falling from rest.
• Moving upwards after being thrown.
• Momentarily stationary at the highest point of its motion.
• Moving downwards after reaching its highest point.

Free fall does not mean that an object must be moving downwards. An object thrown upwards is in free fall as soon as it leaves the thrower’s hand, provided air resistance is ignored.

During its entire flight, gravity accelerates the object downwards.

In a stroboscopic image, the growing gaps between successive positions of a falling object show that its speed is increasing.

Acceleration due to gravity

Near Earth’s surface, all freely falling objects have approximately the same downward acceleration:

g = 9.81 m/s²

For many calculations, this is rounded to:

g = 9.8 m/s²

Some introductory problems use:

g = 10 m/s²

This means that an object’s downward velocity changes by approximately 9.8 m/s every second.

This table assumes that the object begins from rest and air resistance is negligible.

The value of g varies slightly with altitude and location, but 9.8 m/s² is a useful approximation near Earth’s surface.

Mass and free-fall acceleration

In the absence of air resistance, objects of different masses fall with the same acceleration.

A heavy object experiences a larger gravitational force than a light object, but it also has more inertia. These effects balance so that both objects have the same gravitational acceleration.

A bowling ball and a small metal ball dropped together in a vacuum reach the floor at the same time.

On the Moon, where there is almost no atmosphere, an astronaut demonstrated that a hammer and feather fall together. Near the Moon’s surface, the gravitational acceleration is about 1.62 m/s².

Free fall and air resistance

A real object falling through air usually experiences:

• Weight acting downwards.
• Air resistance acting opposite to its motion.

If air resistance is significant, gravity is not the only force acting and the object is not in ideal free fall.

Air resistance depends on factors including:

• Speed.
• Shape.
• Cross-sectional area.
• Air density.

A crumpled piece of paper falls faster than a flat sheet because it experiences less drag relative to its weight.

The constant-acceleration equations using a = g are most accurate when air resistance is negligible.

Choosing a positive direction

Free-fall calculations require a clear sign convention.

Upwards chosen as positive

If upwards is positive:

• Upward displacement is positive.
• Downward displacement is negative.
• Upward velocity is positive.
• Downward velocity is negative.
• Acceleration due to gravity is a = −9.8 m/s².

Downwards chosen as positive

If downwards is positive:

• Downward displacement is positive.
• Downward velocity is positive.
• Acceleration due to gravity is a = +9.8 m/s².

Either convention works. The important requirement is to use one convention consistently throughout the problem.

Upward and downward motion

Consider a ball thrown vertically upwards.

While rising

• Velocity points upwards.
• Acceleration points downwards.
• Velocity and acceleration act in opposite directions.
• The ball slows down.

At the highest point

• Instantaneous velocity is zero.
• Acceleration is still g downwards.
• The ball is about to change direction.

While falling

• Velocity points downwards.
• Acceleration points downwards.
• Velocity and acceleration act in the same direction.
• The ball speeds up.

Gravity does not change direction when the object reaches its highest point.

The position graph reaches its maximum at 2 seconds. The velocity graph has a constant gradient of −g and crosses zero at the same time.

Equations for free fall

Free fall is uniformly accelerated motion, so the standard kinematic equations apply:

v = u + at

s = ½(u + v)t

s = ut + ½at²

v² = u² + 2as

s = vt − ½at²

When upwards is positive, substitute:

a = −g

This gives:

v = u − gt

s = ut − ½gt²

v² = u² − 2gs

The symbols represent:

• s: Vertical displacement.
• u: Initial vertical velocity.
• v: Final vertical velocity.
• a: Vertical acceleration.
• g: Magnitude of gravitational acceleration.
• t: Time.

Use displacement rather than total distance in the vector equations.

Objects that are dropped

The word dropped means that the object begins from rest:

u = 0

It does not mean that acceleration is zero.

For an object dropped from rest:

v = gt

s = ½gt²

when downward is chosen as positive.

Worked example: object dropped from rest

A stone is dropped from a bridge and falls for 3.0 seconds. Ignore air resistance and use g = 9.8 m/s². Calculate its final velocity and displacement.

Choose downwards as positive.

Known:

u = 0 m/s
a = +9.8 m/s²
t = 3.0 s

Final velocity

v = u + at

v = 0 + 9.8(3.0)

v = 29.4 m/s downwards

Displacement

s = ut + ½at²

s = 0 + ½(9.8)(3.0²)

s = 4.9(9.0)

s = 44.1 m downwards

The stone travels farther during each successive second because its velocity is increasing.

Finding fall time from height

Worked example

A ball is dropped from a height of 19.6 m. Calculate the time required to reach the ground. Ignore air resistance.

Choose downwards as positive.

Known:

u = 0 m/s
s = 19.6 m
a = 9.8 m/s²
t = ?

Use:

s = ut + ½at²

19.6 = 0 + ½(9.8)t²

19.6 = 4.9t²

t² = 4

t = ±2

The physical solution is:

t = 2.0 s

The negative solution lies before the selected starting time and does not describe the fall after release.

Finding impact velocity without time

Worked example

A stone is dropped from a height of 45 m. Calculate its impact velocity. Use g = 10 m/s² and ignore air resistance.

Choose downwards as positive.

Known:

u = 0 m/s
s = 45 m
a = 10 m/s²
v = ?

Time is not given, so use:

v² = u² + 2as

v² = 0² + 2(10)(45)

v² = 900

v = 30 m/s

The impact velocity is:

30 m/s downwards

The negative square-root solution is not appropriate for an object falling downwards under this sign convention.

Objects thrown downwards

An object thrown downwards has a non-zero initial downward velocity.

Worked example

A ball is thrown downwards from a building at 5.0 m/s. Find its velocity after 2.0 seconds. Use g = 9.8 m/s².

Choose downwards as positive.

Known:

u = +5.0 m/s
a = +9.8 m/s²
t = 2.0 s

Use:

v = u + at

v = 5.0 + 9.8(2.0)

v = 24.6 m/s downwards

A common mistake is to set u = 0. The phrase “thrown downwards” means the ball already has an initial velocity.

Objects thrown upwards

When an object is thrown upwards, gravity reduces its upward velocity until it reaches zero. The object then begins moving downwards.

Choose upwards as positive:

a = −9.8 m/s²

Worked example: time to maximum height

A ball is thrown upwards at 19.6 m/s. Find the time taken to reach its highest point.

Known:

u = +19.6 m/s
v = 0 m/s
a = −9.8 m/s²
t = ?

Use:

v = u + at

0 = 19.6 − 9.8t

9.8t = 19.6

t = 2.0 s

At the highest point, velocity is zero but acceleration remains −9.8 m/s².

Finding maximum height

Using the same ball:

u = +19.6 m/s
v = 0 m/s
a = −9.8 m/s²
s = ?

Time is not required, so use:

v² = u² + 2as

0² = 19.6² + 2(−9.8)s

0 = 384.16 − 19.6s

19.6s = 384.16

s = 19.6 m

The ball rises 19.6 m above its release point.

Returning to the launch height

If air resistance is ignored and an object returns to the height from which it was launched:

• Time rising equals time falling.
• Speed on returning equals launch speed.
• Return velocity has the opposite direction.
• Total displacement is zero.
• Total distance is twice the maximum height.

For the ball launched upwards at 19.6 m/s:

• Time to rise = 2.0 s.
• Total flight time = 4.0 s.
• Maximum height = 19.6 m.
• Return velocity = −19.6 m/s when upwards is positive.
• Total distance = 39.2 m.
• Displacement = 0 m.

This symmetry applies only when launch and landing heights are equal and air resistance is ignored.

Worked example: launch and landing at the same height

A ball is thrown upwards at 14.7 m/s and returns to its release point. Calculate its total flight time. Ignore air resistance.

Choose upwards as positive.

First find the time to maximum height:

v = u + at

0 = 14.7 − 9.8t

t = 14.7/9.8

t = 1.5 s

The downward journey takes the same amount of time:

Total flight time = 2(1.5)

Total flight time = 3.0 s

Launching from an elevated position

If an object is thrown upwards from a building, its landing position is below its release point. The upward and downward parts are not symmetrical around the release point.

A displacement equation may produce two mathematical times. Interpret each result carefully.

Worked example

A ball is thrown vertically upwards at 15 m/s from a platform 20 m above the ground. Calculate when it reaches the ground. Use g = 10 m/s².

Choose the release point as s = 0 and upwards as positive.

The ground is 20 m below the release point:

s = −20 m

Known:

u = +15 m/s
a = −10 m/s²
s = −20 m

Use:

s = ut + ½at²

−20 = 15t − 5t²

Rearrange:

5t² − 15t − 20 = 0

Divide by 5:

t² − 3t − 4 = 0

Factorize:

(t − 4)(t + 1) = 0

Therefore:

t = 4 s or t = −1 s

The physical solution after release is:

t = 4.0 s

The negative value refers to the mathematical extension of the motion before t = 0.

Determining impact velocity

Continue the platform example:

u = +15 m/s
a = −10 m/s²
t = 4 s

Use:

v = u + at

v = 15 − 10(4)

v = −25 m/s

The impact velocity is:

25 m/s downwards

The negative sign indicates that the velocity points opposite to the chosen positive direction.

Free fall on motion graphs

Position–time graph

For an object thrown upwards, the position–time graph is a downward-opening parabola.

• Positive gradient: moving upwards.
• Zero gradient: at maximum height.
• Negative gradient: moving downwards.
• Increasingly steep negative gradient: downward speed is increasing.

Velocity–time graph

The velocity–time graph is a straight line with gradient −g when upwards is positive.

• Above the axis: upward velocity.
• On the axis: momentarily at rest.
• Below the axis: downward velocity.
• Area under the graph: displacement.

Acceleration–time graph

Acceleration remains constant at −g, so the graph is a horizontal line below the time axis.

Gravity does not become zero at maximum height.

Distance fallen during successive seconds

A dropped object travels increasingly large distances during equal time intervals.

For an object dropped from rest using g = 9.8 m/s²:

The velocity increases by equal amounts each second, while the distance travelled during each second increases.

This occurs because displacement depends on time squared:

s = ½gt²

Reaction time and falling objects

A simple classroom demonstration can estimate reaction time using a falling ruler.

If a ruler falls a measured distance s before being caught:

s = ½gt²

Rearrange:

t = √(2s/g)

Worked example

A ruler falls 0.20 m before being caught.

t = √[2(0.20)/9.8]

t = √0.0408

t ≈ 0.20 s

The estimated reaction time is approximately 0.20 seconds.

The estimate assumes that the ruler begins from rest and falls freely before being caught.

Terminal velocity is not free fall

As an object’s speed through air increases, air resistance usually increases.

Eventually, air resistance may equal the object’s weight:

• Resultant force becomes zero.
• Acceleration becomes zero.
• Velocity becomes constant.

This constant falling speed is called terminal velocity.

An object at terminal velocity is not in ideal free fall because both gravity and air resistance act on it.

A skydiver may be described casually as being in “free fall,” but the ideal physics definition applies only when gravity is the sole force.

A reliable method for free-fall problems

1. Draw a simple vertical diagram.
2. Choose upwards or downwards as positive.
3. List s, u, v, a and t.
4. Give all vector quantities appropriate signs.
5. Replace a with +g or −g according to the sign convention.
6. Select a kinematic equation.
7. Substitute and calculate.
8. State the answer with units and direction.
9. Interpret multiple or negative solutions.
10. Check whether ignoring air resistance is reasonable.

Common misconceptions

• “Free fall means moving downwards.” An upward-moving object can be in free fall.
• “Heavier objects fall faster.” Without air resistance, all objects have the same gravitational acceleration.
• “Acceleration becomes zero at maximum height.” Velocity is zero there; acceleration remains downward.
• “An object that is dropped has zero acceleration.” It has zero initial velocity.
• “Gravity is positive 9.8 m/s² in every problem.” Its sign depends on the chosen positive direction.
• “A negative velocity means an object is slowing down.” It indicates direction.
• “The flight is always symmetrical.” This requires equal launch and landing heights and negligible air resistance.
• “Terminal velocity is free fall.” Air resistance acts at terminal velocity.
• “Every mathematical time is physically meaningful.” Check whether the result lies within the interval being studied.

Did you know?

Astronauts in orbit appear weightless because they and their spacecraft are falling together around Earth.

Gravity is still strong at the altitude of the International Space Station. The astronauts experience apparent weightlessness because no supporting surface pushes against them in the usual way.

Key terms

• Free fall: Motion in which gravity is the only force acting.
• Gravitational acceleration: The acceleration produced by gravity.
• g: The magnitude of gravitational acceleration, approximately 9.8 m/s² near Earth’s surface.
• Initial velocity: Velocity at the beginning of an interval.
• Final velocity: Velocity at the end of an interval.
• Maximum height: The greatest vertical position reached.
• Impact velocity: Velocity immediately before striking a surface.
• Sign convention: A chosen system for representing opposite directions.
• Air resistance: A drag force opposing motion through air.
• Terminal velocity: Constant falling velocity reached when drag balances weight.
• Weight: The gravitational force acting on an object.
• Vacuum: A region containing little or no matter.

Key takeaways

• Free fall occurs when gravity is the only force acting.
• Near Earth’s surface, g is approximately 9.8 m/s² downwards.
• Free fall is uniformly accelerated motion when g is treated as constant.
• A dropped object has u = 0, not a = 0.
• At maximum height, velocity is zero but acceleration remains downward.
• Upward and downward motion require consistent vector signs.
• The standard kinematic equations apply to ideal free fall.
• Equal launch and landing heights produce symmetrical motion when air resistance is ignored.
• Interpret answers using their signs, units and physical context.

4. Analyzing Projectile Paths

Understanding a projectile’s path

A projectile’s motion results from the combination of two perpendicular motions:

• Horizontal motion, which has constant velocity when air resistance is ignored.
• Vertical motion, which has constant downward acceleration due to gravity.

The projectile’s actual velocity is the vector sum of its horizontal and vertical velocity components.

These components can be analyzed independently because gravity acts vertically. The time variable connects the two calculations.

For the ideal model:

aₓ = 0

aᵧ = −g

when upward is chosen as positive.

Near Earth’s surface:

g ≈ 9.8 m/s²

Equations for horizontal motion

With no air resistance, horizontal velocity remains constant:

vₓ = uₓ

Horizontal displacement is:

x = uₓt

The horizontal motion determines how far the projectile travels during its time in the air.

Equations for vertical motion

The vertical motion has constant downward acceleration.

When upward is positive:

vᵧ = uᵧ − gt

y = uᵧt − ½gt²

vᵧ² = uᵧ² − 2gy

These equations use only vertical quantities.

The vertical motion usually determines:

• Time to maximum height.
• Maximum height.
• Total flight time.
• Impact velocity.

Resolving the launch velocity

For a projectile launched at speed u and angle θ above the horizontal:

uₓ = u cos θ

uᵧ = u sin θ

The horizontal component controls the rate of horizontal travel. The vertical component controls the initial upward motion.

A greater launch speed increases both components when the launch angle remains fixed. Changing the angle changes how the initial velocity is divided between horizontal and vertical motion.

Important points on the trajectory

A projectile diagram often identifies three important stages.

Launch

At launch:

• Horizontal velocity is uₓ.
• Vertical velocity is uᵧ.
• Resultant velocity is u.
• Acceleration is g downwards.

Maximum height

At maximum height:

• Vertical velocity is zero.
• Horizontal velocity remains uₓ.
• Acceleration remains g downwards.
• The projectile is momentarily moving horizontally.

Descent and impact

During descent:

• Horizontal velocity remains constant in the ideal model.
• Vertical velocity is negative when upwards is positive.
• Downward speed increases.
• Resultant velocity points forwards and downwards.

The velocity vector is always tangent to the trajectory.

Analyzing trajectory diagrams

The left graph compares launches at 30°, 45° and 60° using the same initial speed of 24 m/s.

The right graph represents a projectile launched at 20 m/s and 35° from 12 m above the ground.

The graphs reveal several patterns:

• The 60° launch reaches a greater height than the 30° launch.
• The 60° launch remains airborne longer.
• The 30° and 60° launches have equal ideal ranges.
• The 45° launch gives the greatest range for equal launch and landing heights.
• An elevated launch produces a longer descent than ascent relative to the release point.
• When launch and landing heights differ, the trajectory is not symmetrical about the launch height.

Finding time to maximum height

At the highest point:

vᵧ = 0

Use:

vᵧ = uᵧ − gt

Therefore:

0 = uᵧ − gt_up

Rearrange:

t_up = uᵧ/g

This result depends only on the initial vertical component.

Worked example

A projectile is launched at 24 m/s at 60° above the horizontal.

Find its initial vertical velocity:

uᵧ = 24 sin 60°

uᵧ ≈ 20.8 m/s

Time to maximum height:

t_up = 20.8/9.8

t_up ≈ 2.12 s

Finding maximum height

Maximum height above the launch point can be found using:

vᵧ² = uᵧ² − 2gH

At maximum height, vᵧ = 0:

0 = uᵧ² − 2gH

Therefore:

H = uᵧ²/(2g)

Using uᵧ = 20.8 m/s:

H = 20.8²/[2(9.8)]

H ≈ 22.0 m

This is the height above the launch point.

If the projectile is launched from an initial height h₀, its maximum height above the ground is:

h_max = h₀ + H

Finding total flight time

Equal launch and landing heights

When a projectile lands at the same height from which it was launched, the ideal vertical motion is symmetrical.

Time falling equals time rising:

T = 2t_up

Therefore:

T = 2uᵧ/g

Different launch and landing heights

If the landing height differs from the launch height, this shortcut cannot be used.

Instead, solve the vertical position equation:

y = y₀ + uᵧt − ½gt²

This normally produces a quadratic equation in t.

Select the solution that represents the time after launch.

Finding horizontal range

The horizontal range is:

R = uₓT

For equal launch and landing heights, substituting the component equations gives:

R = u²sin(2θ)/g

This range formula assumes:

• No air resistance.
• Constant gravitational acceleration.
• Equal launch and landing heights.
• A launch angle measured from the horizontal.

It should not be used for an elevated or lowered landing point.

Worked example: complete equal-height analysis

A ball is launched from ground level at 24 m/s at 30° above the horizontal. Calculate its maximum height, flight time and range. Use g = 9.8 m/s².

Resolve the initial velocity

Horizontal component:

uₓ = 24 cos 30°

uₓ ≈ 20.8 m/s

Vertical component:

uᵧ = 24 sin 30°

uᵧ = 12.0 m/s

Calculate maximum height

H = uᵧ²/(2g)

H = 12.0²/[2(9.8)]

H ≈ 7.35 m

Calculate flight time

T = 2uᵧ/g

T = 2(12.0)/9.8

T ≈ 2.45 s

Calculate range

R = uₓT

R = 20.8(2.45)

R ≈ 50.9 m

The projectile reaches approximately 7.35 m, remains airborne for 2.45 s and travels 50.9 m horizontally.

Comparing complementary angles

Two angles are complementary if they add to 90°.

For example:

30° + 60° = 90°

For the same launch speed and equal launch and landing heights:

R = u²sin(2θ)/g

For θ = 30°:

sin(60°) ≈ 0.866

For θ = 60°:

sin(120°) ≈ 0.866

Therefore, both launch angles produce the same ideal range.

However:

• The 30° trajectory is lower.
• The 30° projectile has greater horizontal velocity.
• The 60° trajectory is higher.
• The 60° projectile remains airborne longer.

Equal range does not mean identical motion.

Why 45° gives the maximum ideal range

For a fixed launch speed:

R = u²sin(2θ)/g

The largest possible value of sin(2θ) is 1.

Therefore:

2θ = 90°

θ = 45°

This conclusion applies to an ideal projectile launched and landing at the same height.

In real sports, the best angle may differ because of:

• Air resistance.
• Spin.
• Release height.
• Object shape.
• The athlete’s ability to produce different speeds at different angles.

Multi-step problem: angled launch from a platform

A ball is launched from a 12 m high platform at 20 m/s and 35° above the horizontal. Calculate:

• Its maximum height above the ground.
• Its time of flight.
• Its horizontal range.
• Its impact velocity.

Use g = 9.8 m/s² and ignore air resistance.

Resolve the launch velocity

uₓ = 20 cos 35°

uₓ ≈ 16.4 m/s

uᵧ = 20 sin 35°

uᵧ ≈ 11.5 m/s

Find the maximum height

First calculate the additional height above the platform:

H = uᵧ²/(2g)

H = 11.5²/[2(9.8)]

H ≈ 6.71 m

Add the platform height:

h_max = 12 + 6.71

h_max ≈ 18.7 m above the ground

Find the time of flight

Use vertical position measured from the launch point. The ground is 12 m below the launch point:

y = −12 m

Use:

y = uᵧt − ½gt²

Substitute:

−12 = 11.5t − 4.9t²

Rearrange:

4.9t² − 11.5t − 12 = 0

Use the quadratic formula:

t = [11.5 ± √(11.5² + 4(4.9)(12))]/9.8

This gives approximately:

t = 3.12 s or t = −0.79 s

The negative time lies before launch, so:

Flight time = 3.12 s

Find the horizontal range

R = uₓT

R = 16.4(3.12)

R ≈ 51.2 m

Find the impact velocity components

Horizontal velocity remains constant:

vₓ = 16.4 m/s

Vertical velocity:

vᵧ = uᵧ − gt

vᵧ = 11.5 − 9.8(3.12)

vᵧ ≈ −19.1 m/s

The negative sign indicates downward motion.

Find the resultant impact speed

v = √(vₓ² + vᵧ²)

v = √(16.4² + 19.1²)

v ≈ 25.2 m/s

Find the impact direction

θ = tan⁻¹(|vᵧ|/vₓ)

θ = tan⁻¹(19.1/16.4)

θ ≈ 49.4°

The impact velocity is approximately:

25.2 m/s at 49.4° below the horizontal

Reading position–time graphs

Projectile motion produces separate horizontal and vertical position graphs.

Horizontal position–time graph

Horizontal position changes linearly:

x = x₀ + uₓt

The graph is a straight line with gradient uₓ.

Vertical position–time graph

Vertical position follows:

y = y₀ + uᵧt − ½gt²

The graph is a downward-opening parabola.

Its gradient represents vertical velocity:

• Positive gradient during ascent.
• Zero gradient at maximum height.
• Negative gradient during descent.

A height–time graph shows height as time changes. It does not show the projectile’s path through space.

Reading velocity–time graphs

Horizontal velocity–time graph

The graph is horizontal because vₓ remains constant.

Its area gives horizontal displacement.

Vertical velocity–time graph

The graph is a straight line with gradient −g:

vᵧ = uᵧ − gt

Its area gives vertical displacement.

• Positive area represents upward displacement.
• Negative area represents downward displacement.
• Equal positive and negative areas give zero total vertical displacement.

If the projectile returns to its launch height, the net area under the vertical velocity graph is zero.

Reading acceleration–time graphs

For the ideal model:

aₓ = 0

The horizontal acceleration graph lies on the time axis.

aᵧ = −g

The vertical acceleration graph is a horizontal line at −9.8 m/s² when upward is positive.

Acceleration does not become zero at maximum height.

Finding position at a particular time

Worked example

A ball is launched at 18 m/s at 40° above the horizontal. Find its position relative to the launch point after 1.5 seconds. Use g = 9.8 m/s².

Resolve the initial velocity:

uₓ = 18 cos 40°

uₓ ≈ 13.8 m/s

uᵧ = 18 sin 40°

uᵧ ≈ 11.6 m/s

Horizontal position:

x = uₓt

x = 13.8(1.5)

x ≈ 20.7 m

Vertical position:

y = uᵧt − ½gt²

y = 11.6(1.5) − 4.9(1.5²)

y ≈ 17.4 − 11.0

y ≈ 6.4 m

After 1.5 seconds, the ball is approximately 20.7 m horizontally from the launch point and 6.4 m above it.

Determining whether a projectile clears an obstacle

A projectile problem may ask whether an object passes over a wall, defender or barrier.

A useful method is:

1. Use horizontal motion to find the time taken to reach the obstacle.
2. Substitute that time into the vertical equation.
3. Compare the projectile’s height with the obstacle’s height.

Worked example

A ball is launched from ground level at 22 m/s at 40°. A 5.0 m wall is located 18 m away. Determine whether the ball clears it. Use g = 9.8 m/s².

Horizontal component:

uₓ = 22 cos 40°

uₓ ≈ 16.9 m/s

Time to reach the wall:

t = x/uₓ

t = 18/16.9

t ≈ 1.07 s

Vertical component:

uᵧ = 22 sin 40°

uᵧ ≈ 14.1 m/s

Height at the wall:

y = uᵧt − ½gt²

y = 14.1(1.07) − 4.9(1.07²)

y ≈ 15.1 − 5.61

y ≈ 9.5 m

Clearance:

9.5 − 5.0 = 4.5 m

The ball clears the wall by approximately 4.5 m under the ideal model.

Factors affecting projectile motion

Initial speed

Increasing launch speed generally increases:

• Maximum height.
• Flight time.
• Horizontal range.
• Impact speed.

For an equal-height launch, range is proportional to speed squared:

R ∝ u²

Doubling the launch speed can produce four times the ideal range when angle and gravity remain unchanged.

Launch angle

The launch angle determines how velocity is divided between horizontal and vertical components.

• A lower angle gives a larger horizontal component.
• A higher angle gives a larger vertical component.
• A 45° angle gives maximum ideal range for equal launch and landing heights.

Launch height

A greater launch height normally increases:

• Time of flight.
• Horizontal range.

It does not change the additional rise produced by the same initial vertical velocity.

Gravitational acceleration

A greater gravitational acceleration causes:

• Shorter flight time.
• Lower maximum height.
• Shorter range.

The same launch would produce a different path on the Moon.

The Moon’s gravitational acceleration is about 1.62 m/s², so a projectile would remain airborne longer and travel farther than on Earth for the same launch conditions.

Air resistance

Air resistance generally:

• Reduces horizontal velocity.
• Reduces range.
• Makes the trajectory less symmetrical.
• Reduces impact speed.
• Changes the path from a perfect parabola.

Drag has a larger effect on objects that are light, large or moving quickly.

Shape and orientation

Streamlined objects usually experience less drag. Irregular or broad objects may slow more quickly.

Spin

Spin can create an additional force called the Magnus force. This can cause a ball to curve, dip or remain airborne differently from the ideal model.

Spin is important in sports such as football, tennis, baseball and golf. A basic no-drag projectile model does not include this effect.

Real-world sports analysis

Projectile analysis can help investigate:

• The best basketball release angle.
• Whether a football will clear a defensive wall.
• The range of a javelin.
• The height of a volleyball serve.
• The landing position of a long jumper’s centre of mass.

Real athletic performance requires models that may include release height, air resistance, spin and movement of the athlete.

A reliable multi-step method

1. Draw and label the trajectory.
2. Choose positive x- and y-directions.
3. Record the launch height.
4. Resolve the initial velocity into uₓ and uᵧ.
5. Create separate horizontal and vertical variable lists.
6. Use vertical motion to determine time.
7. Use the shared time to calculate horizontal range.
8. Find velocity components at the required moment.
9. Recombine components when resultant speed or direction is needed.
10. Check signs, units, assumptions and physical meaning.

Common misconceptions

• “The launch speed is the horizontal velocity.” This is true only for a horizontal launch.
• “Maximum height occurs halfway across every trajectory.” This requires equal launch and landing heights.
• “Flight time is always twice the time to maximum height.” This applies only to equal-height motion.
• “The range formula works for an elevated launch.” The standard formula assumes equal launch and landing heights.
• “A larger launch angle always produces a greater range.” Range reaches a maximum and then decreases.
• “Gravity changes horizontal velocity.” Gravity acts vertically in the ideal model.
• “Velocity is zero at maximum height.” Only the vertical component is zero.
• “A trajectory graph and height–time graph are the same.” They have different horizontal axes.
• “The ideal model exactly predicts sports motion.” Drag, spin and release conditions can cause significant differences.

Did you know?

A computer can predict a projectile’s location by repeatedly updating its velocity and position over very short time intervals.

This numerical approach allows more realistic models to include changing air resistance, wind and spin, even when a simple algebraic equation is unavailable.

Key terms

• Trajectory: The path followed by a projectile.
• Vector component: Part of a vector acting along a selected axis.
• Maximum height: Greatest vertical position reached by a projectile.
• Time of flight: Total time between launch and landing.
• Horizontal range: Horizontal displacement from launch to landing.
• Launch angle: Direction of the initial velocity relative to the horizontal.
• Launch height: Initial vertical position relative to the landing level.
• Impact velocity: Resultant velocity immediately before landing or collision.
• Complementary angles: Two angles whose sum is 90°.
• Air resistance: A force opposing motion through air.
• Magnus force: A force on a spinning object moving through a fluid.
• Ideal model: A simplified representation that ignores complicating effects.

Key takeaways

• Resolve the launch velocity into horizontal and vertical components.
• Horizontal velocity remains constant in the ideal model.
• Vertical acceleration remains equal to g downwards.
• Vertical motion usually determines flight time.
• Horizontal range equals horizontal velocity multiplied by flight time.
• At maximum height, vertical velocity is zero.
• Equal-height trajectories are symmetrical when air resistance is ignored.
• The standard range formula does not apply when launch and landing heights differ.
• Initial speed, launch angle, height, gravity, drag and spin affect the path.
• Multi-step problems require consistent signs and separate horizontal and vertical calculations.

5. Kinematics in Sports and Engineering

Kinematics in Sports and Engineering

Learning targets

• I can identify examples of projectile motion in sports.
• I can explain how engineers use kinematics to design systems and structures.
• I can analyze motion in real-world applications using physics principles.
• I can evaluate how launch angle and speed affect projectile performance.
• I can apply kinematic concepts to practical and technological situations.

Why is kinematics useful?

Kinematics describes motion using quantities such as:

• Position and displacement.
• Distance travelled.
• Speed and velocity.
• Acceleration.
• Time.

In sports, kinematics helps athletes and coaches analyze performance. In engineering, it helps designers predict how machines, vehicles and structures will behave.

Kinematic analysis can answer practical questions such as:

• How high will a ball travel?
• Will a projectile clear an obstacle?
• Where should a conveyor deposit its material?
• How much distance does a vehicle need to stop?
• How quickly should an elevator accelerate?
• When must a robotic arm begin slowing down?
• How far will an object travel after leaving a ramp?

Building a useful model

A real situation must be simplified before equations can be applied.

A kinematic model may assume:

• Motion occurs along a straight line.
• Acceleration remains constant.
• Air resistance is negligible.
• Gravitational acceleration is constant.
• An object can be represented as a single point.
• The surface is level.
• The launch and landing heights are known.

A model does not reproduce every detail of reality. It focuses on the factors needed to answer a particular question.

Projectile motion in sports

Many sports involve projectiles:

• A basketball travelling towards the hoop.
• A football kicked through the air.
• A javelin after release.
• A volleyball serve.
• A golf ball after impact.
• A baseball after being hit or thrown.
• An athlete’s centre of mass during a long jump.

After release, an ideal sports projectile has:

• Constant horizontal velocity.
• Constant downward acceleration.
• A curved parabolic trajectory.

Real sports projectiles are also affected by air resistance, spin, shape and wind. The ideal model provides a useful starting point.

Horizontal and vertical components

For a launch speed u at an angle θ above the horizontal:

uₓ = u cos θ

uᵧ = u sin θ

When air resistance is ignored:

aₓ = 0

aᵧ = −g

The horizontal motion is:

x = uₓt

The vertical motion is:

y = y₀ + uᵧt − ½gt²

Horizontal and vertical calculations share the same time.

How launch angle changes performance

Increasing the launch angle gives more of the initial velocity to the vertical component and less to the horizontal component.

Low launch angle

A low angle produces:

• A large horizontal component.
• A small vertical component.
• A low trajectory.
• A short flight time.
• Possible difficulty clearing obstacles.

High launch angle

A high angle produces:

• A smaller horizontal component.
• A larger vertical component.
• A higher trajectory.
• A longer flight time.
• Greater sensitivity to wind and drag.

Intermediate launch angle

An intermediate angle balances horizontal and vertical motion.

For an ideal projectile launched and landing at the same height:

R = u²sin(2θ)/g

The maximum theoretical range occurs at 45°.

The sports graph compares projectiles launched at the same speed. The engineering graph shows that increasing conveyor speed increases the horizontal landing distance without changing the fall time.

Analyzing the sports graph

Each projectile is launched at 22 m/s from ground level.

A 4 m obstacle is positioned 30 m from the launch point.

• The 25° projectile is below the obstacle and does not clear it.
• The 40° projectile clears it.
• The 55° projectile clears it by a larger vertical distance.
• A higher path is not automatically the best path if range, travel time or accuracy also matters.

The most useful trajectory depends on the performance goal.

Worked example: clearing a defensive wall

A football is kicked at 22 m/s at 40° above the horizontal. A defensive wall 4.0 m high is located 30 m away. Determine whether the ball clears the wall. Ignore air resistance and use g = 9.8 m/s².

Resolve the launch velocity

uₓ = 22 cos 40°

uₓ ≈ 16.9 m/s

uᵧ = 22 sin 40°

uᵧ ≈ 14.1 m/s

Find the time to reach the wall

Horizontal motion has constant velocity:

x = uₓt

t = x/uₓ

t = 30/16.9

t ≈ 1.78 s

Find the ball’s height

y = uᵧt − ½gt²

y = 14.1(1.78) − 4.9(1.78²)

y ≈ 25.1 − 15.5

y ≈ 9.6 m

Find the clearance

Clearance = 9.6 − 4.0

Clearance = 5.6 m

The ideal model predicts that the ball clears the wall by approximately 5.6 m.

The calculation does not determine whether the shot enters the goal. That would require the goal’s position, the ball’s later trajectory and possibly the goalkeeper’s motion.

Launch speed and performance

Launch speed has a strong effect on projectile performance.

For equal launch and landing heights:

R = u²sin(2θ)/g

Therefore:

Range is proportional to the square of launch speed.

If launch speed doubles while the angle remains constant, the ideal range becomes four times as large.

Maximum height also depends on the square of the vertical launch speed:

H = uᵧ²/(2g)

A small increase in launch speed can therefore produce a substantial increase in height and range.

However, faster motion also makes air resistance more significant.

Worked example: effect of increasing launch speed

Two balls are launched at 45° from ground level.

• Ball A is launched at 15 m/s.
• Ball B is launched at 20 m/s.

Use g = 9.8 m/s².

At 45°, sin 90° = 1.

For Ball A:

R = 15²/9.8

R ≈ 23.0 m

For Ball B:

R = 20²/9.8

R ≈ 40.8 m

The speed increases by approximately 33%, but the ideal range increases by approximately 77%.

Basketball trajectory

A basketball shot must satisfy several conditions:

• Reach the horizontal position of the hoop.
• Be high enough to pass over the rim.
• Enter at a suitable downward angle.
• Avoid excessive speed that could cause it to rebound strongly.
• Clear defenders.

A higher release point reduces the required vertical rise. A steeper downward entry can increase the effective opening through the rim, although it may require a higher trajectory.

A real shot is also affected by backspin and air resistance.

Javelin and throwing events

A javelin is not a simple point projectile.

Its flight depends on:

• Release speed.
• Release angle.
• Release height.
• Aerodynamic lift and drag.
• Angle of attack.
• Wind.
• Orientation at landing.

A basic projectile model helps estimate the importance of speed, angle and height. More advanced analysis uses aerodynamic forces and computer simulation.

The ideal 45° maximum-range result does not give the best real javelin release angle because aerodynamic effects and release conditions change the motion.

Long jump and jumping sports

An athlete’s centre of mass behaves approximately like a projectile after take-off.

Coaches may measure:

• Take-off speed.
• Horizontal and vertical velocity components.
• Time in the air.
• Maximum height.
• Landing distance.

The athlete can move their arms and legs during flight, but these movements do not greatly change the projectile path of the whole body’s centre of mass once airborne.

They can, however, change body orientation and help position the feet for landing.

Engineering and projectile motion

Projectile calculations are used when an object leaves a moving system and travels through the air.

Examples include:

• Material leaving a conveyor belt.
• Water leaving a pipe or fountain.
• Packages released from moving equipment.
• Components moving between production stations.
• Emergency supply drops.
• Test objects launched during safety trials.

Engineers need to predict the landing position so that containers, barriers and other equipment can be placed safely.

Conveyor-release model

Suppose material leaves a horizontal conveyor 3.0 m above the floor.

The initial vertical velocity is zero:

uᵧ = 0

The vertical fall time is:

y = ½gt²

t = √(2y/g)

t = √[2(3.0)/9.8]

t ≈ 0.782 s

The fall time is determined by the height and gravity. It does not depend on the horizontal conveyor speed in the ideal model.

Horizontal landing distance is:

x = uₓt

Increasing conveyor speed increases the horizontal distance but does not change the fall time.

Worked engineering problem

A conveyor moves packages horizontally at 3.5 m/s. The packages leave the belt 1.8 m above the centre of a collection bin. Where should the bin be positioned? Ignore air resistance.

Find the fall time

Choose downward as positive.

uᵧ = 0 m/s
y = 1.8 m
aᵧ = 9.8 m/s²

Use:

y = ½aᵧt²

1.8 = 4.9t²

t² = 1.8/4.9

t ≈ 0.606 s

Find the horizontal distance

x = uₓt

x = 3.5(0.606)

x ≈ 2.12 m

The centre of the bin should be approximately 2.12 m horizontally from the conveyor edge under the ideal model.

In practice, engineers would allow for package size, rotation, air resistance and variation in conveyor speed.

Water fountains and fluid streams

A stream of water leaving a nozzle can follow an approximately projectile-shaped path.

Engineers can adjust:

• Nozzle speed.
• Nozzle angle.
• Nozzle height.
• Target landing position.
• Pump pressure.

Individual water droplets follow projectile paths after leaving the nozzle, although drag and interaction between droplets make a real stream more complex.

Transportation engineering

Kinematics is essential in vehicle and road-system design.

Engineers analyze:

• Acceleration lanes.
• Braking distances.
• Railway stopping zones.
• Elevator motion.
• Roller-coaster speeds.
• Vehicle collision tests.
• Runway lengths.
• Traffic-signal timing.

Stopping distance

Total stopping distance consists of:

Thinking distance + braking distance

Thinking distance is:

dₜ = utᵣ

For uniform braking:

v² = u² + 2as

When the vehicle stops, v = 0.

A greater initial speed increases both thinking and braking distances.

Worked example: runway stopping distance

An aircraft touches down at 70 m/s and decelerates uniformly at 3.5 m/s². Estimate the runway distance required to stop.

Choose the direction of motion as positive:

u = 70 m/s
v = 0 m/s
a = −3.5 m/s²

Use:

v² = u² + 2as

0 = 70² + 2(−3.5)s

0 = 4900 − 7s

s = 700 m

The ideal model predicts a braking distance of 700 m.

A real runway must provide additional distance for:

• Pilot response.
• Weather conditions.
• Variable braking.
• Safety margins.
• Aircraft mass and configuration.
• Possible rejected landings or emergencies.

Engineering design uses conservative safety margins rather than treating the ideal calculation as the exact required runway length.

Elevator and ride design

Elevators and amusement rides involve several motion stages:

• Acceleration.
• Constant velocity.
• Deceleration.
• Rest.

Engineers must consider both travel time and passenger comfort.

A very large acceleration can produce uncomfortable forces. A sudden change in acceleration, called jerk, can also be uncomfortable even if the acceleration itself remains within acceptable limits.

Motion controllers vary motor output to create a smooth velocity profile.

Robotics and automated systems

Robotic systems use kinematics to control:

• Position.
• Velocity.
• Acceleration.
• Timing.
• Collision avoidance.
• Tool paths.

A robotic arm must arrive at the correct position with an appropriate velocity. If it moves too quickly, it may overshoot, damage a component or create a safety hazard.

Sensors measure actual motion, and a controller compares the measurements with the planned motion.

Interpreting motion graphs in engineering

Position–time graphs

The gradient gives velocity.

Engineers can use the graph to determine:

• When a system reaches a particular position.
• Whether motion is smooth.
• Whether an object stops or reverses.
• Whether two components may collide.

Velocity–time graphs

The gradient gives acceleration, while area gives displacement.

The graph can show:

• Acceleration and braking stages.
• Maximum operating speed.
• Total travel distance.
• Reversal of direction.

Acceleration–time graphs

Area gives change in velocity.

The graph can reveal:

• Sudden impacts.
• Vibrations.
• Changes in machine operation.
• Passenger or product loads.

Using experimental data

Motion data can be collected using:

• High-speed video.
• Radar.
• GPS.
• Accelerometers.
• Light gates.
• Motion sensors.
• Timing gates.
• Computer vision.

Sports analysts may track a ball frame by frame. Engineers may place sensors on a vehicle, machine or structure.

Measured data can be compared with a mathematical model. Differences may reveal:

• Air resistance.
• Friction.
• Changing acceleration.
• Sensor uncertainty.
• Incorrect assumptions.
• Variation in human performance.

Evaluating a kinematic model

A good analysis does more than produce a numerical answer. It evaluates the model.

Ask:

• Is acceleration reasonably constant?
• Is air resistance significant?
• Are launch and landing heights equal?
• Is the object spinning?
• Are measurements sufficiently precise?
• Does the object have a complex shape?
• Is there wind?
• Does the model need a safety margin?
• Does the calculated result agree with observed motion?

A simple model may still be useful if its assumptions and limitations are stated clearly.

Ideal and real projectile paths

The ideal model is valuable because it isolates the main relationships. More complex models build on it.

Safety factors in engineering

Engineering systems are rarely designed exactly to a theoretical minimum.

A calculated stopping distance of 700 m does not mean that a 700 m runway section is automatically sufficient. Engineers include safety factors to account for variation and uncertainty.

Safety margins may account for:

• Measurement uncertainty.
• Wear and maintenance.
• Weather.
• Material variation.
• Human response.
• Unexpected operating conditions.
• Model limitations.

A practical analysis method

1. Identify the performance or design question.
2. Draw and label the physical situation.
3. Choose coordinate directions.
4. State the assumptions.
5. Identify known and unknown variables.
6. Resolve vectors where necessary.
7. Select suitable equations or graph methods.
8. Calculate the required values.
9. Compare the prediction with the performance target.
10. Evaluate limitations and safety margins.

Common misconceptions

• “The greatest launch angle gives the greatest range.” Very high angles produce long flight times but small horizontal velocities.
• “A 45° angle is always best.” It gives maximum ideal range only for equal launch and landing heights without drag.
• “Increasing launch speed increases range by the same percentage.” Ideal range depends on speed squared.
• “A higher trajectory is always better.” The best path depends on the task.
• “The conveyor speed changes the fall time.” In the ideal model, fall time depends on vertical motion.
• “A calculated engineering value can be used without a safety margin.” Real conditions and uncertainty must be considered.
• “Sports projectiles follow perfect parabolas.” Air resistance and spin alter their paths.
• “Kinematics is used only for projectiles.” It applies to vehicles, elevators, robots and many other moving systems.

Did you know?

A golf ball’s dimples deliberately change the airflow around the ball. They reduce some forms of drag and help spin generate lift, allowing the ball to remain airborne longer than a smooth ball launched under similar conditions.

Its real flight therefore differs substantially from the simplest projectile model.

Key terms

• Sports biomechanics: The application of mechanical principles to human movement and sports.
• Projectile motion: Two-dimensional motion under gravity after launch.
• Launch angle: Direction of initial velocity relative to the horizontal.
• Launch speed: Magnitude of the initial velocity.
• Range: Horizontal displacement from launch to landing.
• Release height: Height of a projectile at launch.
• Trajectory: Path followed by a moving object.
• Kinematic model: A mathematical representation of motion.
• Motion sensor: A device used to measure position or motion.
• Safety factor: Extra design capacity added to account for uncertainty.
• Jerk: Rate of change of acceleration.
• Drag: A resistive force opposing motion through a fluid.
• Magnus force: A force caused by the motion and spin of an object through a fluid.

Key takeaways

• Many sports involve projectile motion after an object is released.
• Launch speed, angle and height affect maximum height, flight time and range.
• The best trajectory depends on the performance goal.
• Engineers use kinematics to predict landing positions, stopping distances and movement times.
• Conveyor-release problems combine horizontal constant velocity with vertical free fall.
• Sports and engineering data can be interpreted using motion graphs.
• Real motion may differ from ideal predictions because of drag, spin and changing acceleration.
• Engineering applications require assumptions, testing and safety margins.
• Kinematics connects mathematical models to practical decisions.