What is projectile motion?

Mixing the horizontal and vertical motions of a projectile. Horizontal velocity is constant; vertical motion has acceleration -g. Keep the two component calculations separate and link them only through time.

A projectile is launched at 30\,m s^-1 horizontally. What is its horizontal velocity after 2\,s? [1 mark]

CCEA CCEA A2 2019-style practice: A ball is projected from ground level with speed 25\,m s^-1 at 40^ above the horizontal. Taking g = 9.8\,m s^-2, find the time of flight and the horizontal range.

Resolve the initial velocity: horizontal u_x = 25cos 40^ = 19.15\,m s^-1, vertical u_y = 25sin 40^ = 16.07\,m s^-1. Vertically (up positive, a = -9.8), the ball returns to the ground when its vertical displacement is zero. Using s = u_y t - (1)/(2)gt^2 = 0: 0 = 16.07t - 4.9t^2 = t(16.07 - 4.9t), so t = 0 or t = 16.074.9 = 3.28\,s. The time of flight is 3.28\,s. The horizontal range is u_x × t = 19.15 × 3.28 = 62.8\,m. Markers reward resolving into components, the time of flight from the vertical motion, and the range from the horizontal motion.

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How do you handle motion with variable acceleration and the flight of a projectile?

Kinematics with variable acceleration using calculus, motion in two dimensions with vectors, and projectile motion treating the horizontal and vertical components separately.

A CCEA A2 Further Maths Mechanics answer on kinematics with variable acceleration using differentiation and integration, motion in two dimensions with position, velocity and acceleration vectors, and projectile motion analysed by resolving into horizontal and vertical components.

Generated by Claude Opus 4.813 min answerUpdated 2026-06-16

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking

CCEA wants you to handle variable acceleration using calculus (differentiating displacement to velocity to acceleration, and integrating the other way), work with motion in two dimensions using vectors, and analyse projectile motion by treating the horizontal and vertical components independently.

The answer

Variable acceleration by calculus

Finding key features of the motion

Motion in two dimensions with vectors

Projectile motion

Variable acceleration and distance

A particle has velocity $v = 3t^2 - 12\,\text{m s}^{-1}$ at time $t$ seconds, moving in a straight line from the origin. Find its acceleration at $t = 2$ and its displacement after $4\,\text{s}$ .

step Acceleration by differentiation

a = \frac{dv}{dt} = 6t, \qquad a(2) = 6(2) = 12\,\text{m s}^{-2}.

step Displacement by integration

s = \int_0^4 (3t^2 - 12)\,dt = \left[t^3 - 12t\right]_0^4 = (64 - 48) - 0 = 16\,\text{m}.

step Interpret

The displacement after $4\,\text{s}$ is $16\,\text{m}$ . (The velocity is negative for $t < 2$ and positive after, so the distance travelled would exceed the displacement; here the net displacement is $16\,\text{m}$ .)

Examples in context

Example 1. A long jumper's trajectory. The athlete's centre of mass follows a projectile path, so the optimum take-off angle balances horizontal speed against airtime. The range formula $\frac{U^2\sin 2\theta}{g}$ shows why a take-off near $45^\circ$ (adjusted for the high launch point) maximises distance.

Example 2. A rocket with changing thrust. While fuel burns, a rocket's acceleration changes with time, so engineers integrate the time-dependent acceleration to get velocity and then position. This is the variable-acceleration calculus model rather than suvat.

Try this

Q1. A particle has displacement $s = 2t^3$ . Find its velocity at $t = 1$ . [2 marks]

Cue. $v = \frac{ds}{dt} = 6t^2 = 6\,\text{m s}^{-1}$ .

Q2. A projectile is launched at $30\,\text{m s}^{-1}$ horizontally. What is its horizontal velocity after $2\,\text{s}$ ? [1 mark]

Cue. Still $30\,\text{m s}^{-1}$ (horizontal velocity is constant).

Q3. For variable acceleration, how do you find displacement from velocity? [1 mark]

Cue. Integrate the velocity with respect to time.

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 A2 20217 marksA particle moves in a straight line so that its displacement is

s = t^3 - 6t^2 + 9t

metres at time

t

seconds. Find the times when the particle is momentarily at rest, and its acceleration at each of those times.

Show worked answer →

The velocity is the derivative of displacement: $v = \dfrac{ds}{dt} = 3t^2 - 12t + 9$ .

At rest, $v = 0$ : $3t^2 - 12t + 9 = 0$ , so $t^2 - 4t + 3 = 0$ , giving $(t - 1)(t - 3) = 0$ , hence $t = 1\,\text{s}$ or $t = 3\,\text{s}$ .

The acceleration is the derivative of velocity: $a = \dfrac{dv}{dt} = 6t - 12$ .

At $t = 1$ : $a = 6(1) - 12 = -6\,\text{m s}^{-2}$ . At $t = 3$ : $a = 6(3) - 12 = 6\,\text{m s}^{-2}$ .

Markers reward differentiating to get velocity and acceleration, solving the velocity equation, and the two acceleration values.

CCEA A2 20198 marksA ball is projected from ground level with speed

25\,\text{m s}^{-1}

40^\circ

above the horizontal. Taking

g = 9.8\,\text{m s}^{-2}

, find the time of flight and the horizontal range.

Show worked answer →

Resolve the initial velocity: horizontal $u_x = 25\cos 40^\circ = 19.15\,\text{m s}^{-1}$ , vertical $u_y = 25\sin 40^\circ = 16.07\,\text{m s}^{-1}$ .

Vertically (up positive, $a = -9.8$ ), the ball returns to the ground when its vertical displacement is zero. Using $s = u_y t - \frac{1}{2}gt^2 = 0$ :

$0 = 16.07t - 4.9t^2 = t(16.07 - 4.9t),$ so $t = 0$ or $t = \dfrac{16.07}{4.9} = 3.28\,\text{s}.$

The time of flight is $3.28\,\text{s}$ . The horizontal range is $u_x \times t = 19.15 \times 3.28 = 62.8\,\text{m}$ .

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

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Sources & how we know this

CCEA GCE Further Mathematics specification — CCEA (2018)