Projectile Motion A Level: SUVAT, Range and Time of Flight

What is projectile motion?

A projectile is any object that is moving through the air with gravity as the only force acting on it. A kicked football, a thrown ball and a stone dropped from a bridge all behave like projectiles, so long as you can ignore air resistance. The path it follows is a parabola.

The one idea that unlocks every projectile question is independence of motion. Gravity acts vertically, so it changes only the vertical component of the velocity. There is no horizontal force, so the horizontal component of the velocity does not change. Gravity cannot tell how fast the object is moving sideways. A ball dropped from rest and a ball fired horizontally from the same height hit the ground at the same time.

You do two jobs in every question. First, split the motion into a horizontal part and a vertical part. Second, solve each with the right equation and link them using time, the only quantity the two parts share. For other Module 3 topics, see the Module 3 hub.

SUVAT equations and resolving velocity

The SUVAT equations describe motion in a straight line with constant acceleration. The five quantities are displacement s, initial velocity u, final velocity v, acceleration a and time t. Each equation leaves one of them out.

For a projectile launched at speed u at an angle θ above the horizontal, resolve the velocity into components.

Choose a sign convention at the start and stick to it. If you take up as positive, then a = −9.81 m s⁻², and a ball thrown upwards has a positive uᵧ. If you take down as positive, a = +9.81 m s⁻² and an upward launch has a negative uᵧ. Either works, but mixing them is the commonest source of wrong answers.

Horizontal launch

Here the projectile starts with a purely horizontal velocity, for example a ball rolling off a table or a package released from a level-flying plane. The vertical launch component is zero, so the vertical motion is the same as free fall from rest.

• Time of fall: use h = ½gt² (since uᵧ = 0), so t = √(2h/g).
• Horizontal distance: x = uₓt, using that same t.
• Speed on landing: find vᵧ = gt, then combine with uₓ using Pythagoras: v = √(uₓ² + vᵧ²).
• Direction on landing: tan φ = vᵧ/uₓ, where φ is the angle below the horizontal.

The time of fall depends only on the height, not on the horizontal speed. Doubling the launch speed doubles the distance travelled but leaves the time unchanged.

Angled launch: time of flight, maximum height and range

For a launch at an angle θ, the vertical velocity starts at uᵧ = u sin θ, falls steadily at 9.81 m s⁻¹ every second, is zero at the top, and reverses on the way down. For a projectile that lands at the same height it started, the motion is symmetrical.

These results all follow from the SUVAT equations, and you should be able to derive each of them rather than memorise them.

• Time to the top: at the highest point vᵧ = 0, so 0 = u sin θ − gt, giving t = u sin θ / g.
• Time of flight (level ground): the motion is symmetrical, so T = 2u sin θ / g. (Or set y = 0 in y = uᵧt − ½gt² and take the non-zero solution.)
• Maximum height: use v² = u² + 2as with v = 0, so H = u² sin²θ / 2g.
• Range (level ground): R = uₓT = u cos θ × 2u sin θ / g = u² sin 2θ / g, using 2 sin θ cos θ = sin 2θ.

The range is largest when sin 2θ = 1, so θ = 45°. Because sin 2θ = sin (180° − 2θ), angles that add up to 90° (complementary angles) give the same range, for example 30° and 60°, or 20° and 70°.

On a vertical velocity-time graph the gradient is the acceleration (−g) and the area between the line and the time axis is the vertical displacement. The area above the axis equals the area below it when the projectile lands at the same height as it was launched.

The effect of air resistance

Everything above assumes no air resistance. In reality drag acts opposite to the velocity and grows with speed, so it changes the motion in both directions. At A level you need to describe the effects qualitatively.

• The horizontal velocity is no longer constant: drag decelerates the projectile horizontally, so it decreases throughout the flight.
• On the way up, drag and weight both act downwards, so the deceleration is greater than g.
• On the way down, drag acts upwards and opposes weight, so the acceleration is less than g.
• The maximum height is lower and the range is shorter than without air resistance.
• The path is no longer symmetrical: the descent is steeper than the ascent, and the time spent falling is longer than the time spent rising.
• The speed on landing is less than the launch speed, because work has been done against drag.
• The angle for greatest range is a little below 45°.

A light, fast object such as a shuttlecock shows these effects strongly. A dense compact object such as a shot put shows them very little, which is why we can usually ignore air resistance for it.

Practical: measuring g by free fall

This is a standard practical for the motion topic. A steel ball is held by an electromagnet at a measured height above a trapdoor. Switching off the electromagnet starts an electronic timer and the ball falls. The timer stops when the ball hits the trapdoor, so you measure the time t to fall a height s.

1. Measure the height s from the bottom of the ball to the trapdoor with a metre rule (or use a light gate to find a speed).
2. Release the ball and record the time of fall t. Repeat several times at the same height and take the mean.
3. Change the height and repeat for at least six heights.
4. The ball starts from rest, so s = ½gt². Plot s against t². The gradient is g/2, so g = 2 × gradient.

Plotting a graph uses all of the results and helps to show up systematic errors. A line that does not pass through the origin shows a systematic error, such as a delay in the timer or the ball sticking to the magnet. With a single height, you can use g = 2s/t².

Worked examples

Common mistakes

• Using the launch speed in a vertical equation. Use only the vertical component, uᵧ = u sin θ, for vertical motion.
• Saying the velocity is zero at the top. Only the vertical component is zero. The projectile still moves horizontally at uₓ, so its speed at the top is u cos θ.
• Giving the horizontal part an acceleration. Without air resistance, the horizontal acceleration is zero.
• Mixing signs. Decide which direction is positive and apply it to u, v, a and s.
• Using the range formula when the heights differ. R = u² sin 2θ / g is for level ground. Otherwise solve for time first.
• Using 45° for the best angle in every case. It is the best angle only for level ground and no air resistance.
• Thinking mass matters. Mass does not appear in the free-fall equations, so without air resistance a heavy ball and a light ball fall together.

How OCR A examines projectile motion

Projectile motion is in Module 3 (Forces and motion), within the motion topic. Expect these styles.

• Multi-step calculations that start from a horizontal or angled launch and ask for time, range, height or landing velocity.
• Graph questions: sketch or interpret displacement-time or velocity-time graphs for the horizontal and vertical components.
• Explain questions: why a ball dropped and a ball fired horizontally land together, or the effect of air resistance on the path.
• Practical questions on measuring g by free fall: choosing measurements, plotting s against t², calculating g from the gradient, and analysing uncertainties.

Practical work on g falls under the practical activity groups (PAGs), so be ready to describe the method, evaluate it and combine percentage uncertainties. The topic links to momentum and impulse through Newton's laws, to circular motion as another motion with a force acting at right angles to the velocity, and to gravitational fields, where g is the field strength near Earth's surface.

Practice questions

1. A plane flies horizontally at 80 m s⁻¹ at a height of 500 m. A package is released. Ignore air resistance. Calculate how long it takes to reach the ground and how far it travels horizontally in that time.4 marks

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   Answer: 10.1 s and 808 m

   t = √(2h/g) = √(1000 / 9.81) = 10.10 s. x = 80 × 10.10 = 808 m.

2. A cannonball is fired at 30 m s⁻¹ at 60° above the horizontal from level ground. Calculate the maximum height, the time to reach it and the range.5 marks

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   Answer: Height 34.4 m, time 2.65 s, range 79.5 m

   uᵧ = 30 sin 60° = 25.98 m s⁻¹. H = uᵧ²/2g = 675 / 19.62 = 34.4 m. t = uᵧ/g = 25.98 / 9.81 = 2.65 s. R = u² sin 2θ / g = 900 × sin 120° / 9.81 = 79.5 m.

3. A ball is thrown at 18 m s⁻¹ at 25° above the horizontal on level ground. Calculate the range. State another launch angle with the same speed that gives the same range.3 marks

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   Answer: 25.3 m; the same range is given at 65°

   R = u² sin 2θ / g = 324 × sin 50° / 9.81 = 25.3 m. The complementary angle is 90° − 25° = 65°.

4. In a free-fall experiment a ball falls 1.20 m (± 0.001 m) in a mean time of 0.495 s (± 0.005 s). Calculate g and its percentage uncertainty, and state the absolute uncertainty in g.4 marks

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   Answer: g = 9.79 m s⁻², with a 2.1% uncertainty, so g = 9.79 ± 0.21 m s⁻²

   g = 2s/t² = 2 × 1.20 / 0.495² = 9.79 m s⁻². Percentage uncertainty = 0.08% (s) + 2 × 1.0% (t) = 2.1%. Absolute uncertainty = 2.1% of 9.79 = 0.21 m s⁻².

5. Explain how air resistance changes the path of a projectile compared with the path in a vacuum.4 marks

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   Answer: The range and maximum height are smaller; the path is asymmetric, with a steeper descent than ascent; the horizontal velocity decreases during the flight; the landing speed is less than the launch speed.

   Reasons: drag acts opposite to the velocity, so it decelerates the horizontal motion. On the way up, drag adds to weight so the deceleration exceeds g; on the way down, drag opposes weight so the acceleration is less than g.

Next steps

Practise resolving velocity on a few launches until it feels automatic, then try questions that start from a height. Projectile motion leads naturally into momentum and impulse and circular motion. Use the Module 3 hub for the full list. To practise this topic with step-by-step marking, start free.

Related guides

Module 3 · Newton's laws of motion and momentumMomentum and Impulse: Collisions, Explosions and Force-Time Graphs

Module 5 · Circular motionCentripetal Force and Acceleration: Circular Motion Explained

Module 5 · Gravitational fieldsGravitational Field Strength and Potential: Equations, Graphs and Examples

Module 3 · MaterialsYoung Modulus Experiment: Stress, Strain and the Wire Practical