Projectiles
A projectile is any object thrown or launched that then moves under gravity alone. The trick is to split the motion into two independent parts: steady horizontal motion and vertical motion with constant acceleration.
Part 1 of 3: Learn it
In short
- Horizontally there is no acceleration, so horizontal distance is horizontal speed times time.
- Vertically: acceleration g downwards, so use suvat.
- Time links the two directions: it is the same for both.
Where this is in your specification
Spec points: DfE Q5 (AQA 7357 Q5, Edexcel 9MA0 Mechanics topic 7, OCR A H240 3.02)
| Board | Topic: Projectiles |
|---|---|
| DfE content | Q5 |
| AQA 7357 | Q5 |
| Edexcel 9MA0 | Mechanics topic 7 (7.5) |
| OCR H240 | 3.02 |
Splitting the velocity
Launched at speed u and angle θ above the horizontal, the components are u cos θ horizontally and u sin θ vertically. Questions may give tan θ as a fraction; then build a right-angled triangle to get sin θ and cos θ exactly.
Two directions, one time
| Horizontal | Vertical (up positive) | |
|---|---|---|
| Initial velocity | u cos θ | u sin θ |
| Acceleration | 0 | −g |
| Displacement | x = (u cos θ)t | y = (u sin θ)t − ½gt² |
Useful results
- At the greatest height the vertical velocity is 0; the horizontal velocity is unchanged.
- On level ground, the time of flight is 2u sin θ ÷ g.
- Speed at any moment: combine the two components with Pythagoras.
The model ignores air resistance and treats the object as a particle. Real objects fall short of the predicted range.
What is the horizontal acceleration of a projectile in this model?
Show the answer
Zero.
Part 2 of 3: See it worked
Worked examples
Example 1
A ball is kicked at 20 m/s at 30° above level ground. Find the time of flight, the range and the greatest height. (g = 9.8 m/s²)
- Components: 20 cos 30° = 17.32 m/s and 20 sin 30° = 10 m/s
- Time: 0 = 10t − 4.9t², so t = 10 ÷ 4.9 = 2.041 s
- Range: 17.32 × 2.041 = 35.35 m
- Height: 0 = 10² − 2 × 9.8 × h, so h = 100 ÷ 19.6 = 5.10 m
Answer: 2.04 s, 35.3 m and 5.10 m (3 s.f.).
Example 2
A stone is thrown horizontally at 12 m/s from the top of a 19.6 m cliff. How far from the foot of the cliff does it land?
- Vertical: u = 0, a = 9.8 down, s = 19.6: 19.6 = 4.9t², so t² = 4 and t = 2 s
- Horizontal: 12 × 2
Answer: 24 m.
Common mistakes
- Using g in the horizontal direction.
- Mixing up sin and cos when resolving: the vertical part uses sin θ when θ is measured from the horizontal.
- Saying the velocity at the top is zero. Only the vertical part is zero.
- Using the full speed u in a vertical suvat equation.
A ball is projected at 15 m/s at 40°. Write down its vertical component.
Show the answer
15 sin 40° = 9.64 m/s (3 s.f.).
Part 3 of 3: Test yourself
Check yourself
Answer each one in your head or on paper first, then open it to check.
What is the horizontal acceleration of a projectile in this model?
Zero.
A ball is projected at 15 m/s at 40°. Write down its vertical component.
15 sin 40° = 9.64 m/s (3 s.f.).
Give one assumption made in the projectile model.
Any one of: no air resistance, the object is a particle, g is constant.
Jobs that use this
Each link opens the job profile on the National Careers Service (England). In the rest of the UK: My World of Work (Scotland), Careers Wales, nidirect careers (Northern Ireland).
Full lessons and marked practice for this course are coming soon to Brainlag Learn. See courses