Projectile Motion Explained: A-Level Physics (9702)

Projectile motion is two separate motions happening at once. Horizontally the velocity is constant, because no horizontal force acts. Vertically the object accelerates under gravity. Resolve the launch velocity into horizontal and vertical components, apply the equations of uniformly accelerated motion to each direction, and share only the time between them.

The idea behind everything

A projectile here means an object moving only under gravity after launch, with air resistance ignored. The 9702 syllabus asks you to describe and explain motion due to a uniform velocity in one direction and a uniform acceleration in a perpendicular direction, and that is exactly this.

• Horizontal: no force acts, so the velocity is constant.
• Vertical: gravity acts, so the acceleration is g = 9.81 \text{ m s}^{-2} downwards.

Time is the only quantity the two directions share. Once you find the time of flight, you can find range, height or final velocity.

Splitting the launch velocity

If an object is launched at speed u at angle \theta above the horizontal, the components are:

• Horizontal: u_x = u\cos\theta
• Vertical: u_y = u\sin\theta

The horizontal component never changes. The vertical component decreases as the object rises, is zero at the top, then increases downwards.

The equations you need

These are the equations for uniformly accelerated motion, applied to each direction separately. Taking up as positive for the vertical direction:

Two results save time for a launch and landing at the same height:

• Time to reach maximum height: t = \frac{u\sin\theta}{g}
• Maximum height: H = \frac{(u\sin\theta)^2}{2g}

Worked example 1: launched at an angle

A ball is launched at u = 20 \text{ m s}^{-1} at 30^\circ above the horizontal. Take g = 9.81 \text{ m s}^{-2} and ignore air resistance. This is an example in the style of an exam question.

Step 1: resolve the velocity.

u_x = 20\cos 30^\circ = 17.3 \text{ m s}^{-1}

u_y = 20\sin 30^\circ = 10.0 \text{ m s}^{-1}

Step 2: time to maximum height. The vertical velocity is zero at the top.

t = \frac{u_y}{g} = \frac{10.0}{9.81} = 1.02 \text{ s}

Step 3: maximum height.

H = \frac{u_y^2}{2g} = \frac{10.0^2}{2 \times 9.81} = 5.10 \text{ m}

Step 4: total time of flight. For a level launch and landing, the time up equals the time down.

T = 2t = 2.04 \text{ s}

Step 5: horizontal range. The horizontal velocity is constant.

R = u_x \times T = 17.3 \times 2.04 = 35.3 \text{ m}

Worked example 2: thrown horizontally off a cliff

A stone is thrown horizontally at 12 \text{ m s}^{-1} from a cliff 45 \text{ m} above the sea. Take g = 9.81 \text{ m s}^{-2} and ignore air resistance.

Time to fall. Vertically the initial velocity is zero, so s = \frac{1}{2}gt^2:

t = \sqrt{\frac{2 \times 45}{9.81}} = 3.03 \text{ s}

Horizontal distance from the base of the cliff.

x = 12 \times 3.03 = 36.4 \text{ m}

Velocity as it hits the sea. The vertical component is v_y = gt = 9.81 \times 3.03 = 29.7 \text{ m s}^{-1} downwards, and the horizontal component is still 12 \text{ m s}^{-1}. The speed is

v = \sqrt{12^2 + 29.7^2} = 32.0 \text{ m s}^{-1}

and the angle below the horizontal is \tan^{-1}\left(\frac{29.7}{12}\right) = 68.0^\circ.

The pattern is the same as before: find the time from the vertical motion, then use it for the horizontal motion.

Sign conventions that catch people out

• Pick a positive direction for vertical motion and stick to it. If up is positive, g is negative in your equations.
• Never put g into a horizontal calculation. Horizontal motion uses constant velocity.
• The speed at any moment is v = \sqrt{v_x^2 + v_y^2}, and its direction is given by \tan\theta = \frac{v_y}{v_x}.
• Check the launch and landing heights. The shortcut T = 2t only works when they are equal.

A resolving checklist

Before you touch an equation, write down what you know in each direction. Under "horizontal" put the initial horizontal velocity and note that acceleration is zero. Under "vertical" put the initial vertical velocity, the acceleration g with its sign, and any displacement given, such as the height of a cliff. Most errors happen because a value from one column is used in the other.

Common mistakes

• Using the total speed in a directional equation. The launch speed u is not the vertical or horizontal velocity. Always use the resolved component in the matching direction.
• Assuming symmetry when it does not hold. The time up equals the time down only when launch and landing are at the same height. From a cliff, the object lands lower than it started, so you must solve the vertical displacement equation for the whole flight instead of doubling the time to the top.
• Forgetting that the quadratic has two roots. When you solve y = u_y t - \frac{1}{2}gt^2 for time, reject the root that is negative or that belongs to a different part of the motion.

How the marks are usually given

Mark schemes generally split a projectile question into stages, so each stage can earn credit separately: a mark for resolving the velocity correctly, a mark for a suitable equation for time, a mark for using that time in the other direction, and a mark for the final value with its unit. If you make a slip in an early stage, a later stage can still be credited when you carry your own value forward correctly, which is called follow-through. The exact allocation varies between questions, so check the mark scheme.

Air resistance

The syllabus asks you to describe and explain this qualitatively, so it is an explanation question rather than a calculation. With air resistance:

• The range is shorter and the maximum height is lower.
• The horizontal velocity is no longer constant, because a drag force acts against the motion.
• The path is no longer symmetrical, and the descent is steeper than the climb.

A good explanation links cause and effect: drag acts opposite to the velocity, so the horizontal component of velocity decreases, so less distance is covered. How to answer explain questions in 9702 shows how to turn this into marks.

Quick checklist

• Resolve into u\cos\theta and u\sin\theta.
• Horizontal velocity is constant, and vertical velocity changes by g each second.
• Find the time from one direction and use it in the other.
• At maximum height the vertical component is zero, not the whole velocity.
• Quote answers to a sensible number of significant figures, with units.

Practise projectile questions on real papers

Projectile questions are a good test of method, because the steps are marked as well as the answer. Open the A Level Physics past papers, write your working by hand and have it marked against the official mark scheme, with follow-through marking applied when an early value is wrong. Reading and downloading papers is free, and a free account adds AI marking and Pal, which gives hints but not answers. The 2023 and 2024 papers are marked first, with earlier years being added.