Topic 4 of 8 · Form 5
Projectiles
A particle moving under gravity only (no air resistance), with constant acceleration g downwards.
Notes for this topic 1 file, in order
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Assumptions
A particle moving under gravity only (no air resistance), with constant acceleration g downwards.
Equations
For a projection with speed u at angle θ above the horizontal:
- Horizontal: x = u cos θ · t (constant velocity).
- Vertical: y = u sin θ · t − ½gt².
- Time of flight (level ground) T = 2u sin θ/g; maximum height H = u² sin²θ/(2g); range R = u² sin 2θ/g (maximum at θ = 45°).
- Equation of trajectory: y = x tan θ − gx²/(2u² cos²θ) = x tan θ − gx²(1 + tan²θ)/(2u²).
Worked example: A ball is kicked at 20 m s⁻¹ at 30° to the horizontal from level ground. Find the time of flight, range and greatest height (g = 9.8).
T = 2 × 20 × 0.5 / 9.8 = 2.04 s
R = 400 × sin 60° / 9.8 = 35.3 m
H = 400 × 0.25 / 19.6 = 5.10 m
Hitting a target
Substitute the target's coordinates into the trajectory equation and solve the quadratic in tan θ – there are usually two possible angles.
Velocity at any time
vx = u cos θ; vy = u sin θ − gt; speed = √(vx² + vy²); direction tan⁻¹(vy/vx).
Check yourself
- At what angle is the range maximum on level ground?
- A stone is thrown horizontally at 12 m s⁻¹ from a height of 19.6 m. Find the horizontal distance travelled (g = 9.8).
- What is the vertical velocity at the highest point?
Show answers
- 45°
- t = 2 s → 24 m
- Zero.
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