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How do we analyse projectile motion in two dimensions and motion with a varying acceleration?

Projectile motion resolved into horizontal and vertical components, the range, time of flight and maximum height, and using calculus to relate displacement, velocity and acceleration when the acceleration varies with time.

A CCEA A-Level Mathematics answer on projectile motion resolved into horizontal and vertical components, finding the time of flight, range and maximum height, and using differentiation and integration to relate displacement, velocity and acceleration when acceleration varies with time.

Generated by Claude Opus 4.813 min answer

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What this dot point is asking

CCEA wants you to analyse projectile motion by resolving it into independent horizontal and vertical components, find the time of flight, range and maximum height, and use calculus to relate displacement, velocity and acceleration when the acceleration is not constant. This extends the straight-line kinematics of AS 2 into two dimensions and into variable motion.

The answer

Projectiles: two independent motions

Time of flight, range and maximum height

The range is greatest when sin⁡2α=1\sin 2\alpha = 1, that is at α=45∘\alpha = 45^\circ. In an exam these formulae are usually rebuilt from the components rather than quoted, so the reliable method is to resolve the initial velocity, find the time of flight from the vertical motion, and then use that time in the horizontal motion to find the range.

Variable acceleration and calculus

Worked example: distance from a variable velocity

Examples in context

Example 1. The angle for maximum range. A long jumper or a thrown ball achieves the greatest horizontal range (from level ground, no air resistance) at a launch angle of 45∘45^\circ, where sin⁡2α=1\sin 2\alpha = 1. The range formula explains this familiar result directly.

Example 2. Braking with a variable force. A vehicle whose braking force changes has a time-dependent acceleration, so its stopping distance is found by integrating the velocity, not by suvat. Calculus kinematics handles the realistic case where acceleration is not constant.

Try this

Q1. A projectile is launched at 14 m s−114\,\text{m s}^{-1} at 30∘30^\circ. Find the vertical component of velocity. [1 mark]

  • Cue. 14sin⁡30∘=7 m s−114\sin 30^\circ = 7\,\text{m s}^{-1}.

Q2. A particle has displacement s=t3−2ts = t^3 - 2t. Find its velocity. [2 marks]

  • Cue. v=dsdt=3t2−2v = \frac{ds}{dt} = 3t^2 - 2.

Q3. At what launch angle is the range of a projectile greatest (from level ground)? [1 mark]

  • Cue. 45∘45^\circ.

Exam-style practice questions

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

CCEA 20207 marksA ball is projected from ground level at 20 m s−120\,\text{m s}^{-1} at 30∘30^\circ above the horizontal. Taking g=9.8 m s−2g = 9.8\,\text{m s}^{-2}, find the time of flight and the horizontal range.
Show worked answer →

Resolve the initial velocity: horizontal ux=20cos⁡30∘=17.32 m s−1u_x = 20\cos 30^\circ = 17.32\,\text{m s}^{-1}, vertical uy=20sin⁡30∘=10 m s−1u_y = 20\sin 30^\circ = 10\,\text{m s}^{-1}.

Vertical motion (up positive, a=−9.8a = -9.8): the ball returns to the ground when y=0y = 0. Using y=uyt−12gt2y = u_y t - \frac{1}{2}gt^2:

0=10t−4.9t2=t(10−4.9t),0 = 10t - 4.9t^2 = t(10 - 4.9t), so t=104.9=2.04 st = \frac{10}{4.9} = 2.04\,\text{s}.

Horizontal range: x=uxt=17.32×2.04=35.3 mx = u_x t = 17.32 \times 2.04 = 35.3\,\text{m}.

Markers reward resolving the velocity, the time of flight from the vertical motion, and the range from the horizontal motion.

CCEA 20196 marksA particle moves in a straight line so that its velocity is v=3t2−12t+9v = 3t^2 - 12t + 9 metres per second at time tt seconds. Find the times when it is instantaneously at rest, and its acceleration when t=1t = 1.
Show worked answer →

At rest when v=0v = 0: 3t2−12t+9=03t^2 - 12t + 9 = 0, so t2−4t+3=0t^2 - 4t + 3 = 0, giving (t−1)(t−3)=0(t - 1)(t - 3) = 0 and t=1 st = 1\,\text{s} or t=3 st = 3\,\text{s}.

The acceleration is a=dvdt=6t−12a = \frac{dv}{dt} = 6t - 12.

At t=1t = 1: a=6(1)−12=−6 m s−2a = 6(1) - 12 = -6\,\text{m s}^{-2}.

Markers reward setting v=0v = 0, both times at rest, differentiating for acceleration, and the value at t=1t = 1.

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