OxfordAQA International AS and A-level Physics: Projectile motion, friction, drag and terminal speed (9630)

This study guide follows the OxfordAQA International AS and A-level Physics (9630) specification, version 4.4, section 3.2.4 (Projectile motion). It is shared International AS and A-level content, so it applies whichever route you are taking. Gravitational field strength is taken as 9.81 N kg⁻¹ (m s⁻² for free fall) throughout; in an exam, use the value the question gives you.

Next: the revision notes and the practice questions. The hub is OxfordAQA A-level Physics, the printable list is the checklist, and you can test yourself with the free 10-minute diagnostics. The equations used here come from motion along a straight line and Newton’s laws.

What the specification asks for

Specification wording	What you must be able to do
Independent effect of motion in horizontal and vertical directions of a uniform gravitational field	Treat the two directions separately; problems are solvable with the equations of uniform acceleration
Qualitative treatment of friction	Describe where friction acts and what it does; no calculations. Static and dynamic friction are not tested as separate ideas
Qualitative treatment of lift and drag	Describe the direction and cause of each force and use them in force-balance arguments
Terminal speed	Explain why a falling object reaches a constant speed
Air resistance increases with speed	Use this fact to explain changes in acceleration
Effect of air resistance on trajectory, and on maximum speed of a vehicle	Describe qualitatively how the path changes and which factors set a vehicle’s top speed

Independent horizontal and vertical motion

A projectile is an object moving through the air with only gravity acting on it. Gravity acts vertically downwards, so it changes the vertical velocity and has no effect on the horizontal velocity. The two directions are independent: they share only the time.

Direction	Resultant force	Acceleration	Result
Horizontal	zero	zero	constant horizontal velocity
Vertical	weight	9.81 m s⁻² downwards	uniform acceleration

Method:

1. Choose a sign convention (usually up is positive, so a = −9.81 m s⁻² vertically).
2. Resolve the launch velocity: horizontal component u cos θ, vertical component u sin θ (θ measured from the horizontal).
3. Write the horizontal equation: distance = horizontal velocity × time.
4. Write the vertical equations from v = u + at, s = ½(u + v)t, s = ut + ½at² and v² = u² + 2as.
5. Link the two with the time t.

At the highest point the vertical velocity is zero but the acceleration is still 9.81 m s⁻² downwards. The speed there is not zero: it equals the horizontal component.

Worked example 1: horizontal launch

A pebble is flicked horizontally at 9.5 m s⁻¹ from a footbridge 22.0 m above a river. Find the time of flight, the horizontal distance travelled and the velocity on entering the water.

Vertical: u = 0, s = 22.0 m (down taken as positive here), a = 9.81 m s⁻².

s = ½at², so t = √(2s/a) = √(2 × 22.0 / 9.81) = 2.12 s

Horizontal distance = 9.5 × 2.118 = 20.1 m

Vertical speed on entry: v = at = 9.81 × 2.118 = 20.8 m s⁻¹ downwards.

Speed = √(9.5² + 20.8²) = 22.8 m s⁻¹

Angle below horizontal = tan⁻¹(20.8 / 9.5) = 65.4°

Worked example 2: launch at an angle from the ground

A beanbag leaves the ground at 14.0 m s⁻¹ at 52° above the horizontal and lands at the same height. Find the maximum height, the time of flight and the range.

Components: horizontal 14.0 cos 52° = 8.62 m s⁻¹; vertical 14.0 sin 52° = 11.03 m s⁻¹.

Maximum height: v = 0 at the top, so v² = u² + 2as gives 0 = 11.03² − 2 × 9.81 × H, so H = 6.20 m

Time to the top: t = 11.03 / 9.81 = 1.125 s. Landing at the same height, the descent takes as long as the ascent, so the time of flight is 2.25 s.

Range = 8.62 × 2.249 = 19.4 m

Worked example 3: landing on a higher level

A ball is thrown at 15.0 m s⁻¹ at 60° above the horizontal towards a platform 3.0 m above the launch point. Find when and where it lands.

Components: horizontal 7.50 m s⁻¹; vertical 12.99 m s⁻¹.

Vertical displacement 3.0 m: s = ut + ½at², so 3.0 = 12.99t − 4.905t².

Rearrange: 4.905t² − 12.99t + 3.0 = 0. The formula for a quadratic gives t = 0.256 s or t = 2.39 s.

The ball passes the height of the platform twice: once rising (0.256 s) and once falling (2.39 s). It lands on the platform on the way down, so t = 2.39 s.

Check: the vertical velocity then is 12.99 − 9.81 × 2.393 = −10.5 m s⁻¹, negative, so the ball is falling.

Horizontal distance = 7.50 × 2.393 = 17.9 m

Friction

Friction is a contact force that acts along the surfaces, opposing the relative motion of the surfaces (or the tendency of one surface to slide over the other). It acts on both objects, in opposite directions, as a Newton’s third law pair.

• Its direction is parallel to the surface and opposite to the motion or the way the object would otherwise slide.
• It reduces the resultant force on a moving object and so reduces its acceleration.
• The work done against it transfers energy to internal (thermal) energy of the surfaces.
• Rougher surfaces and larger forces pressing the surfaces together generally mean more friction.

The specification asks only for this qualitative picture. You do not need to separate static from dynamic friction.

Drag and lift

When an object moves through a fluid (a gas or a liquid), the fluid exerts a force on it. Resolve this force into two parts:

• Drag points against the direction in which the object moves through the fluid. In air it is called air resistance.
• Lift acts perpendicular to the direction of the fluid flow past the object. An aerofoil produces lift by deflecting air downwards, so the air pushes the wing upwards (Newton’s third law).

Drag increases with speed. It also depends on the object’s shape and cross-sectional area and on the density of the fluid. You only need the qualitative statement that faster means more drag.

Force balance for an aircraft in level flight at constant velocity: lift equals weight, and thrust equals drag. If thrust is increased, the aircraft accelerates until drag has grown to match the new thrust.

Terminal speed

Consider an object released from rest in air.

1. At the instant of release its speed is zero, so drag is zero. The only force is weight, so the acceleration is g.
2. As it speeds up, drag grows. The resultant force (weight minus drag) falls, so the acceleration falls. The speed is still increasing.
3. When drag equals weight the resultant force is zero. The acceleration is zero and the object falls at a constant velocity: the terminal speed.

A velocity–time graph starts with gradient g and the curve flattens until it becomes horizontal at the terminal speed. If drag is larger at the same speed (for example a parachute opens), the object decelerates until drag again equals weight at a lower terminal speed.

Worked example 4: force balance for a skydiver

A skydiver of mass 80 kg has weight 80 × 9.81 = 785 N. Find her acceleration when the air resistance is 500 N, and state the air resistance at terminal speed.

Resultant force downwards = 785 − 500 = 285 N.

a = F / m = 285 / 80 = 3.56 m s⁻² downwards.

At terminal speed the resultant force is zero, so the air resistance is 785 N.

Effect of air resistance on a projectile’s path

With no air resistance, the path is a symmetrical parabola. With air resistance acting opposite to the velocity:

• The horizontal velocity decreases continuously, because there is now a horizontal force.
• The vertical deceleration on the way up is greater than g, and the vertical acceleration on the way down is less than g.
• The maximum height and range are smaller than the values without air resistance.
• The path is no longer symmetrical: the descent is steeper than the ascent.
• For a projectile returning to its launch height, the landing speed is less than the launch speed, because work has been done against air resistance.

Maximum speed of a vehicle

A vehicle accelerates while the driving force is greater than the total resistive force. The resistive forces include air resistance, which increases with speed, and rolling resistance and friction in the moving parts. The maximum speed is reached when the driving force equals the total resistive force, so the resultant force is zero.

Factors that change the maximum speed:

• The driving force available from the engine: a larger force, or more power, raises the maximum speed.
• Shape: a streamlined shape reduces drag at a given speed.
• Cross-sectional area: a smaller frontal area reduces drag.
• Fluid density: thinner air at altitude means less drag.
• Extra resistance, such as a roof box, lowers the maximum speed if the engine is unchanged.

Worked example 5: resultant force on a car

A car of mass 1100 kg is pushed forward by 1800 N while resistive forces total 1200 N at a certain speed. Find its acceleration, and say what happens as the speed rises.

Resultant force = 1800 − 1200 = 600 N, so a = 600 / 1100 = 0.545 m s⁻².

As speed rises, resistive forces grow. The resultant force and the acceleration fall. When the resistive forces reach 1800 N the acceleration is zero and the car holds its maximum speed.

Common errors

• Using the launch speed in a vertical equation without first resolving it.
• Saying the acceleration is zero at the top of the flight. Only the vertical velocity is zero.
• Adding the horizontal and vertical times together. There is one time, shared by both directions.
• Ignoring the second root of a quadratic: choose the root that matches the situation (falling onto a higher level, or after launch).
• Saying that at terminal speed “no forces act”. The forces are balanced, so the resultant is zero.
• Saying drag stays constant as speed rises. It increases.
• Describing the path with air resistance as symmetrical.

Practise with the practice questions and recall with the revision notes.

Official syllabus

OxfordAQA International AS and A-level Physics (9630) specification, version 4.4, for exams May/June 2020 onwards, published by OxfordAQA. Section 3.2.4, Projectile motion.