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EnglandMathsSyllabus dot point

How do you analyse the motion of an object launched into the air under gravity alone?

Modelling projectile motion by resolving into independent horizontal and vertical components, using the constant acceleration equations, and finding range, maximum height, time of flight and the equation of the path.

A focused answer to the AQA A-Level Mathematics projectiles content, covering resolving motion into independent horizontal and vertical parts, using the suvat equations, and finding range, maximum height, time of flight and the path equation.

Generated by Claude Opus 4.811 min answer

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  1. What this dot point is asking
  2. Resolving the launch
  3. Key quantities
  4. The equation of the path
  5. A general method and the modelling assumptions

What this dot point is asking

AQA wants you to model projectile motion by treating the horizontal and vertical components separately, apply the constant acceleration equations to each, and find quantities such as range, maximum height, time of flight and the equation of the path. The single big idea is independence of the two components, which turns one hard two-dimensional problem into two straightforward one-dimensional ones.

Resolving the launch

The two components are linked only by the shared time tt. You set up a suvat treatment for each component using the same tt, then solve. Time is therefore the bridge between horizontal and vertical motion, and most projectile problems are solved by finding tt first.

Key quantities

For maximum height, use vy=0v_y = 0 at the top with vy2=(usinθ)22gHv_y^2 = (u\sin\theta)^2 - 2gH. For the impact speed when the projectile lands, combine the horizontal and vertical velocity components with Pythagoras, since speed is the magnitude of the velocity vector.

The equation of the path

Eliminating time between the horizontal equation x=(ucosθ)tx = (u\cos\theta)t and the vertical equation y=(usinθ)t12gt2y = (u\sin\theta)t - \frac{1}{2}gt^2 gives yy as a quadratic in xx, which is a parabola. Substituting t=xucosθt = \frac{x}{u\cos\theta} yields y=xtanθgx22u2cos2θy = x\tan\theta - \frac{g x^2}{2u^2\cos^2\theta}. This path equation lets you find the height at a given horizontal distance, or test whether the projectile clears an obstacle.

A general method and the modelling assumptions

Every projectile problem follows the same plan. First, resolve the launch velocity into horizontal and vertical components, ucosθu\cos\theta and usinθu\sin\theta. Second, set up suvat for the vertical motion (with acceleration g-g if up is positive) and note the horizontal motion is at constant velocity. Third, use the shared time tt to link the two: find tt from one component, then use it in the other. Fourth, answer the question, whether that is a height, a range, a velocity at impact, or a position at a given time.

The model rests on assumptions you should be ready to state and criticise: the projectile is a particle (so its size and spin are ignored), air resistance is negligible (so the only force is gravity and the horizontal velocity stays constant), and gg is constant over the flight. Air resistance is the most significant omission: in reality it reduces both the range and the maximum height, and makes the descent steeper than the ascent, so the true path is not a symmetric parabola. Examiners reward candidates who can say how a result would change if air resistance were included.

Exam-style practice questions

Practice questions written in the style of AQA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AQA 20198 marksPaper 2, Section B. A ball is projected from ground level with speed 2525 metres per second at an angle of 4040 degrees above the horizontal. Take g=9.8g = 9.8 metres per second squared and ignore air resistance. (a) Find the time of flight. (b) Find the horizontal range. (c) Find the maximum height reached.
Show worked answer →

Resolve the launch: horizontal ux=25cos4019.15u_x = 25\cos 40^\circ \approx 19.15 m/s (constant); vertical uy=25sin4016.07u_y = 25\sin 40^\circ \approx 16.07 m/s. For (a), the ball returns to ground when vertical displacement is zero: using s=uyt12gt2=0s = u_y t - \frac{1}{2}gt^2 = 0 gives t=2uyg=2×16.079.83.28t = \frac{2u_y}{g} = \frac{2 \times 16.07}{9.8} \approx 3.28 s. For (b), range =ux×t=19.15×3.2862.8= u_x \times t = 19.15 \times 3.28 \approx 62.8 m. For (c), at the top vertical velocity is zero: 0=uy22gH0 = u_y^2 - 2gH, so H=16.0722×9.813.2H = \frac{16.07^2}{2 \times 9.8} \approx 13.2 m. Markers reward resolving the launch, using zero vertical displacement for the flight time, and vy=0v_y = 0 for the maximum height.

AQA 20216 marksPaper 2, Section A. A stone is thrown horizontally with speed 1212 metres per second from the top of a cliff 3030 m above the sea. Take g=9.8g = 9.8 metres per second squared. (a) Find the time taken to reach the sea. (b) Find the horizontal distance from the base of the cliff to where the stone lands. (c) Find the speed of the stone as it hits the sea.
Show worked answer →

Horizontal velocity is constant at 1212 m/s; initial vertical velocity is 00. For (a), the vertical drop is 30=12gt2=4.9t230 = \frac{1}{2}gt^2 = 4.9t^2, so t2=6.122t^2 = 6.122 and t2.47t \approx 2.47 s. For (b), horizontal distance =12×2.4729.7= 12 \times 2.47 \approx 29.7 m. For (c), vertical velocity at impact is vy=gt=9.8×2.4724.2v_y = gt = 9.8 \times 2.47 \approx 24.2 m/s; the speed is 122+24.22144+585.627.0\sqrt{12^2 + 24.2^2} \approx \sqrt{144 + 585.6} \approx 27.0 m/s. Markers reward using zero initial vertical velocity, the drop equation for the time, and combining components with Pythagoras for the impact speed.

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