Projectile Motion — OCR A Level Physics
Module 3 · Forces and Motion

Projectile Motion

Specification: OCR A H556  |  Section: 3.1.3 Projectile motion  |  Focus: splitting motion into horizontal and vertical parts, using components clearly, and solving OCR-style problems with confidence

By the end of this topic you should be able to…
  • split a launch velocity into horizontal and vertical components
  • explain why horizontal and vertical motion are independent
  • solve projectile questions involving time, range and maximum height
  • use simple, clear sign conventions in vertical motion
  • spot common exam traps in multi-step projectile questions

Big idea: a projectile is really two motions happening at once — horizontal motion with constant velocity, and vertical motion with constant acceleration.

Part 1

What is projectile motion?

A projectile is any object moving under the effect of gravity alone after it is launched. In OCR A questions, we usually ignore air resistance unless told otherwise.

This gives two simple rules:

Horizontal motion

No horizontal force means no horizontal acceleration, so horizontal velocity stays constant.

Vertical motion

Gravity acts vertically, so the projectile has constant vertical acceleration equal to g.

Overall path

The combination gives a curved path called a parabola.

OCR wording to remember

The horizontal and vertical components of a projectile’s motion are independent. The time taken in one direction still matters in the other, but the accelerations are treated separately.

Part 2

Resolving the launch speed into components

If a projectile is launched with speed u at angle θ above the horizontal, then:

Horizontal componentux = u cos θ
Vertical componentuy = u sin θ

Once you have these components, treat the two directions separately.

Common mistake

Do not use the full launch speed u in vertical SUVAT equations unless the launch is vertical. Use the vertical component u sin θ instead.

Part 3

Useful projectile-motion ideas

Horizontal motion

Horizontal displacementx = ux t = u cos θ · t

Vertical motion

Vertical displacementy = uy t − 1/2 gt2
Vertical velocityvy = uy − gt

At the highest point

The vertical velocity is zero, but the horizontal velocity is still unchanged.

Maximum height shortcut0 = uy2 − 2gh    so    h = uy2 / 2g

Time of flight

If the projectile lands at the same vertical height from which it was launched:

Time of flight for same launch and landing heightT = 2u sin θ / g

Range

Again, for launch and landing at the same height:

Range formulaR = u cos θ × T = u2 sin 2θ / g
Important condition

The shortcut formulas for time of flight and range only work directly when the projectile lands at the same height as it started.

Interactive

Projectile motion simulator

Launch angle, speed and path
20
40°
Watch the horizontal and vertical motion happen together. The dashed path is the full trajectory, and the moving ball shows the live motion. Try changing angle while keeping speed fixed.
ux: 15.3 m s⁻¹
uy: 12.9 m s⁻¹
Max height: 8.5 m
Range: 40.2 m
Time of flight: 2.63 s
Part 4

Common misconceptions and exam traps

“The horizontal velocity decreases”

Not in the standard OCR model. If air resistance is ignored, horizontal velocity stays constant.

“Acceleration becomes zero at the top”

No. The vertical velocity is zero at the top, but acceleration is still g downward.

“Use one big SUVAT equation for both directions”

No. Split the motion into horizontal and vertical parts.

“Range formula always works”

Only if the launch height and landing height are the same.

High-value exam trap

If the projectile is launched from a cliff or lands on a platform at a different height, you usually need to solve the vertical motion for time first, then use the horizontal motion separately.

Worked Examples

Worked examples

Worked example 1
A ball is launched horizontally at 6.0 m s−1 from a table 1.25 m high. Find a) the time to hit the floor, b) the horizontal distance travelled.
Vertical motion: uy = 0, s = 1.25 m, a = 9.81 m s−2
Use s = ut + 1/2 at2: 1.25 = 0 + 1/2 × 9.81 × t2
t = 0.505 s
Horizontal motion: x = ux t = 6.0 × 0.505 = 3.03 m
Worked example 2
A projectile is launched at 20 m s−1 at 30° to the horizontal. Find a) ux, b) uy.
ux = u cos θ = 20 cos 30° = 17.3 m s−1
uy = u sin θ = 20 sin 30° = 10.0 m s−1
Worked example 3
A projectile is launched at 18 m s−1 at 40°. It lands at the same height. Find the time of flight.
Use T = 2u sin θ / g
T = 2 × 18 × sin 40° / 9.81
T = 2.36 s
Worked example 4
A projectile is launched at 18 m s−1 at 40°. It lands at the same height. Find the range.
Use R = u2 sin 2θ / g
R = 182 × sin 80° / 9.81
R = 32.5 m
Worked example 5
A projectile is launched at 24 m s−1 at 50°. Find the maximum height reached.
Find vertical launch speed: uy = 24 sin 50° = 18.4 m s−1
At the top, vy = 0
Use vy2 = uy2 − 2gh
h = uy2 / 2g = 18.42 / (2 × 9.81) = 17.3 m
Worked example 6
A ball is kicked from ground level at 16 m s−1 at 35°. Find its horizontal speed and explain whether that changes during flight.
ux = 16 cos 35° = 13.1 m s−1
With air resistance ignored, horizontal acceleration is zero
So the horizontal speed stays constant throughout the flight
Worked example 7
A projectile is launched from the edge of a 20 m cliff with horizontal speed 12 m s−1. Find the time to hit the ground and the horizontal distance from the cliff base.
Vertical motion: 20 = 1/2 × 9.81 × t2
t = 2.02 s
Horizontal distance = 12 × 2.02 = 24.2 m
Worked example 8
A projectile is launched at 25 m s−1 at 30° and lands on a platform 5.0 m above the launch point. Find the time it takes to reach the platform.
Vertical motion only: uy = 25 sin 30° = 12.5 m s−1
Use s = ut − 1/2 gt2: 5.0 = 12.5t − 4.905t2
This gives two times: t = 0.47 s on the way up, or 2.08 s on the way down
Always think about which time the question means
Knowledge

Knowledge Check

1
What is the horizontal acceleration of a projectile if air resistance is ignored?
1 mark
  • Zero
2
At the highest point of a projectile’s path, what is a) the vertical velocity and b) the vertical acceleration?
2 marks
  • a) Vertical velocity = 0
  • b) Vertical acceleration = g downward / −9.81 m s−2 if upward is positive
3
Why do two projectiles dropped from the same height hit the ground at the same time if one also has horizontal speed?
2 marks
  • Because horizontal and vertical motion are independent
  • They have the same vertical motion
4
For a projectile launched and landing at the same height, what launch angle gives the maximum range?
1 mark
  • 45°
Exam Practice

Exam-Style Questions

1
Explain what is meant by saying that the horizontal and vertical motions of a projectile are independent.
3 marks
  • Horizontal motion is unaffected by the vertical motion
  • Vertical motion is affected by gravity
  • The two components can therefore be analysed separately
2
A ball is projected horizontally at 8.0 m s−1 from a cliff 4.9 m high.

a) Calculate the time taken to hit the ground. [2 marks]
b) Calculate the horizontal distance from the base of the cliff to the landing point. [2 marks]
c) Calculate the vertical speed just before impact. [2 marks]
6 marks
  • a) 4.9 = 1/2 × 9.81 × t2, so t = 1.00 s
  • b) x = 8.0 × 1.00 = 8.0 m
  • c) vy = gt = 9.81 × 1.00 = 9.81 m s−1 downward
3
A projectile is launched at 22 m s−1 at an angle of 35° to the horizontal. It lands at the same height.

a) Calculate the horizontal component of velocity. [1 mark]
b) Calculate the vertical component of velocity. [1 mark]
c) Calculate the time of flight. [2 marks]
d) Calculate the range. [2 marks]
6 marks
  • a) ux = 22 cos 35° = 18.0 m s−1
  • b) uy = 22 sin 35° = 12.6 m s−1
  • c) T = 2u sin θ / g = 2 × 22 sin 35° / 9.81 = 2.57 s
  • d) R = uxT = 18.0 × 2.57 = 46.3 m
4
A student says, “At the top of the path, the projectile stops, so its acceleration is zero.” Explain why this is wrong.
3 marks
  • Only the vertical velocity is zero at the highest point
  • The projectile still has horizontal velocity
  • Acceleration remains g downward throughout the motion
5
A football is kicked from ground level at 28 m s−1 at 42° to the horizontal.

a) Calculate the maximum height. [3 marks]
b) Calculate the time to reach maximum height. [2 marks]
c) Calculate the total time of flight. [1 mark]
6 marks
  • a) uy = 28 sin 42° = 18.7 m s−1
  • Use h = uy2 / 2g = 18.72 / (2 × 9.81)
  • h = 17.8 m
  • b) t = uy / g = 18.7 / 9.81 = 1.91 s
  • c) Total time = 2 × 1.91 = 3.82 s
6
A projectile is fired from a point 12 m above level ground with speed 18 m s−1 at 25° above the horizontal.

a) Show that the vertical component of launch speed is about 7.6 m s−1. [1 mark]
b) Find the time taken to hit the ground. [4 marks]
c) Find the horizontal range. [2 marks]
7 marks
  • a) uy = 18 sin 25° = 7.61 m s−1
  • b) Take upward as positive: s = −12, u = 7.61, a = −9.81
  • Use s = ut + 1/2 at2: −12 = 7.61t − 4.905t2
  • Solve to get positive root t = 2.43 s
  • c) ux = 18 cos 25° = 16.3 m s−1
  • Range = 16.3 × 2.43 = 39.6 m
7
Two projectiles are launched from the same point with the same speed. One is launched at 30° and the other at 60° to the horizontal. They land at the same height from which they were launched.

a) State what is the same about their ranges. [1 mark]
b) Explain your answer. [2 marks]
c) State one important difference between their flights. [1 mark]
4 marks
  • a) Their ranges are the same
  • b) Because R = u2 sin 2θ / g and sin 60° = sin 120°
  • c) The 60° launch has greater maximum height and longer time of flight
8
A student measures the horizontal distance travelled by a projectile but forgets to convert the launch angle from degrees conceptually into components and simply substitutes the full speed into both horizontal and vertical equations. Explain why this method is wrong.
3 marks
  • The launch speed must be resolved into horizontal and vertical components
  • The horizontal motion depends on u cos θ and the vertical motion depends on u sin θ
  • Using the full speed in both directions double-counts the motion and gives incorrect answers

Topic Summary

Split the motion

Projectile motion works best when you separate horizontal and vertical parts immediately.

Horizontal

No horizontal acceleration in the standard model, so horizontal velocity is constant.

Vertical

Acceleration is always g downward, even at the highest point.

Shortcuts

Range and time-of-flight shortcuts only apply directly when launch and landing height are the same.

ux = u cos θ
uy = u sin θ
x = u cos θ · t
y = u sin θ · t − 1/2 gt2
T = 2u sin θ / g
R = u2 sin 2θ / g