OxfordAQA International AS and A-level Physics: Projectiles and terminal speed (9630) – Revision Notes

These notes condense section 3.2.4 (Projectile motion) of the OxfordAQA International AS and A-level Physics (9630) specification, version 4.4. It is shared International AS and A-level content. For full explanations and longer worked examples read the study guide, then try the practice questions. The course page is OxfordAQA A-level Physics, the checklist lets you tick off outcomes, and the free 10-minute diagnostics show where to focus. The equations come from motion along a straight line and Newton’s laws, with practice on those equations.

The big idea

Horizontal and vertical motion are independent. They share only the time. Gravity affects only the vertical motion.

Direction	Acceleration	Motion
Horizontal (no air resistance)	0	constant velocity
Vertical	9.81 m s⁻² downwards	uniform acceleration

Formulas and what each one is for

Use	Equation
Horizontal distance	x = u_x t, where u_x = u cos θ
Vertical component at launch	u_y = u sin θ
Vertical velocity	v = u + at
Vertical displacement	s = ut + ½at²
Vertical velocity without time	v² = u² + 2as
Average velocity form	s = ½(u + v)t
Speed at any moment	speed = √(v_x² + v_y²)
Direction at any moment	tan(angle) = v_y / v_x

For a launch from and back to the same level (no air resistance):

• Time to the top = u_y / g.
• Time of flight = 2u_y / g.
• Maximum height = u_y² / 2g.
• Range = u_x × time of flight.

Method in steps

1. Draw the path and mark the launch point, landing point and axes.
2. Pick a sign convention and keep it. Up positive means a = −9.81 m s⁻².
3. Resolve the launch velocity into u_x and u_y.
4. Decide what the problem gives you in the vertical direction (s, u, v, a, t). Pick the equation that contains the unknown and nothing you do not know.
5. Use the time in the horizontal equation.
6. Check the answer is sensible: the sign of v_y shows whether the object is rising or falling.

Small worked reminders

Horizontal launch: a pellet leaves a table-top launcher horizontally at 3.5 m s⁻¹, 1.25 m above the floor.

• Vertical: u = 0, s = 1.25 m. t = √(2 × 1.25 / 9.81) = 0.505 s.
• Horizontal distance = 3.5 × 0.505 = 1.77 m.
• Vertical speed on landing = 9.81 × 0.505 = 4.95 m s⁻¹.

Angled launch: a ball leaves the ground at 18.0 m s⁻¹ at 35° above the horizontal.

• u_x = 14.7 m s⁻¹ and u_y = 10.3 m s⁻¹.
• Maximum height = 10.32² / (2 × 9.81) = 5.43 m.
• Time of flight = 2 × 10.32 / 9.81 = 2.10 s.
• Range = 14.74 × 2.105 = 31.0 m.

Launch from a height with an upward component: a ball leaves the top of a 20 m cliff with a horizontal component of 6.0 m s⁻¹ and an upward component of 10.0 m s⁻¹. It lands at the foot of the cliff.

• Take up as positive: s = −20 m, u = 10.0 m s⁻¹, a = −9.81 m s⁻².
• s = ut + ½at² gives 4.905t² − 10.0t − 20 = 0.
• t = (10.0 + √(100 + 392.4)) / 9.81 = 3.28 s. The negative root is rejected because the time must be positive.
• Horizontal distance from the foot of the cliff = 6.0 × 3.28 = 19.7 m.
• Vertical velocity on landing = 10.0 − 9.81 × 3.28 = −22.2 m s⁻¹, so the speed is √(6.0² + 22.2²) = 23.0 m s⁻¹.

Reading graphs of projectile and falling motion

Graph	What it shows
Vertical velocity against time, no air resistance	straight line of gradient −9.81 m s⁻² (up positive), crossing zero at the top
Horizontal velocity against time, no air resistance	horizontal line
Horizontal velocity against time, with air resistance	line that slopes downwards, flattening as the force falls
Velocity against time for a falling object	curve starting with gradient g and levelling out at the terminal speed
Height against horizontal distance	parabola without air resistance; asymmetrical with air resistance

The gradient of a velocity–time graph is the acceleration. The area under it is the displacement. At terminal speed the gradient is zero. Between release and terminal speed the area under the curve is less than the area under a straight line of gradient g, because the acceleration is smaller than g.

Forces: friction, lift and drag

Force	Direction	Key point
Friction	along the surfaces, opposing relative motion or its tendency	reduces acceleration; transfers energy to internal energy; qualitative only
Drag (air resistance)	opposite to the velocity relative to the fluid	increases with speed
Lift	perpendicular to the flow of fluid past the object	aerofoil deflects air downwards, so air pushes the wing up

Static and dynamic friction are not tested as separate topics.

Level flight at constant velocity: lift = weight and thrust = drag.

Terminal speed in steps

1. Released from rest: speed is zero, drag is zero, acceleration is g.
2. Speed rises, so drag rises. Resultant force and acceleration fall.
3. Drag equals weight: resultant force is zero, acceleration is zero.
4. The object falls at a constant terminal speed.

Graph of velocity against time: starts with gradient g, the gradient decreases, and the curve levels off horizontally.

Opening a parachute increases drag at once. Drag exceeds weight, so the skydiver decelerates until drag equals weight again at a lower terminal speed.

Air resistance and a projectile

• Horizontal velocity decreases.
• Maximum height and range are smaller.
• The path is asymmetrical, with a steeper descent than ascent.
• Landing speed at the launch level is less than launch speed.

Maximum speed of a vehicle

Maximum speed occurs when driving force equals total resistive force. Resistive forces rise with speed. A higher maximum speed needs a larger driving force or less resistance (streamlined shape, smaller frontal area). A roof box increases resistance and lowers the maximum speed if the engine is unchanged.

Must-know distinctions

• Zero vertical velocity at the top is not zero acceleration: a is still 9.81 m s⁻² downwards.
• Speed at the top equals the horizontal component, not zero.
• Terminal speed means balanced forces, not no forces.
• Drag acts against the velocity. Lift acts across the flow.
• Without air resistance the path is symmetrical. With it, the path is not.

Quick self-test

Take g = 9.81 m s⁻². Answer without looking, then check.

1. A coin is dropped from rest from the top of a 38 m tower. How long does it take to reach the ground?
2. A pellet is fired horizontally at 8.0 m s⁻¹ from 1.8 m above a floor. How far from the launch point does it land?
3. A projectile leaves with a vertical component of 30 m s⁻¹. Find the time to the highest point and the maximum height above the launch level.
4. A ball thrown horizontally at 5.0 m s⁻¹ has been in flight for 0.60 s. Find its speed.
5. A 2.0 kg object falls with a drag force of 12 N. Find its acceleration.
6. What is the vertical component of the velocity of a projectile at the top of its path, and what is the acceleration there?
7. State the condition for a falling object to reach terminal speed.
8. Name the direction of lift relative to the airflow.
9. Give two ways air resistance changes the path of a projectile.
10. A car has maximum speed on a flat road. State what is true of the driving force and the resistive forces.

Answers

1. t = √(2 × 38 / 9.81) = 2.78 s.
2. t = √(2 × 1.8 / 9.81) = 0.606 s. Distance = 8.0 × 0.606 = 4.85 m.
3. t = 30 / 9.81 = 3.06 s. H = 30² / (2 × 9.81) = 45.9 m.
4. v_y = 9.81 × 0.60 = 5.886 m s⁻¹. Speed = √(5.0² + 5.886²) = 7.72 m s⁻¹.
5. Weight = 2.0 × 9.81 = 19.62 N. Resultant = 19.62 − 12 = 7.62 N. a = 7.62 / 2.0 = 3.81 m s⁻² downwards.
6. The vertical velocity is zero. The acceleration is 9.81 m s⁻² downwards.
7. The drag (and any other upward force) equals the weight, so the resultant force is zero.
8. Perpendicular to the direction of the flow of air past the object.
9. Any two: horizontal velocity falls; range is smaller; maximum height is smaller; descent is steeper; landing speed is lower.
10. The driving force equals the total resistive force, so the resultant force is zero.

Where marks are usually lost

• Not resolving the launch velocity before using an equation.
• Mixing signs: taking up as positive for u and then using +9.81.
• Stopping after the time and not using it to find the distance or speed that the question asks for.
• Reporting only the vertical speed when the speed on landing is asked for. Combine the two components.
• Choosing the wrong root of a quadratic. Check the sign of the vertical velocity.
• Saying “the forces are zero” at terminal speed when the resultant force is zero.
• Describing a graph without saying what the gradient represents.
• Writing “air resistance slows it down” without saying which component or which effect on the path.
• Forgetting units, or giving too many significant figures when the data has two or three.

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.