OxfordAQA International AS and A-level Mathematics: Projectile motion (9660)

This guide covers section M2.5 (Projectiles) of the OxfordAQA International AS and A-level Mathematics (9660) specification, Version 5.2 (International AS exams from May/June 2018, A-level from May/June 2019). The specification states this content is part of International A-level (A2) Unit M2 (Mechanics) and is not examined at International AS.

Related pages: revision notes, practice questions, straight-line kinematics, Newton’s laws and variable acceleration, exam preparation, the course hub, the printable checklist and the free 10-minute diagnostics.

What you must be able to do

Spec	Outcome	Where
M2.5	Describe the motion of a particle under gravity in two dimensions; state and use x = (V cos α)t and y = (V sin α)t − ½gt²	Sections 1 and 2
M2.5	Be aware of the assumptions you are making	Section 1
M2.5	Calculate range, time of flight and maximum height	Sections 3 and 4
M2.5	Eliminate time to derive the equation of the trajectory	Section 5

The specification says the formulae for range, time of flight and maximum height should not be quoted in exams, and that you do not need to know the trajectory formula but must be able to derive it when needed. So the working below always starts from the two equations of motion. Inclined planes and air resistance are not set.

1. The model and its assumptions

A projectile is a particle launched with speed V at angle α above the horizontal. After launch the only force is gravity. Take g = 9.8 m/s² unless a question gives another value.

State these assumptions when asked:

• The object is a particle, so its size, shape and spin are ignored.
• There is no air resistance (resistance problems are not set).
• The acceleration due to gravity is constant, 9.8 m/s² downwards, so the model suits motion near the ground.
• The motion happens in one vertical plane, and the ground is flat unless told otherwise.
• Wind has no effect.

A real ball has air resistance. Resistance slows it down, so a real ball would normally travel a shorter horizontal distance and reach a lower height than the model predicts.

2. Components and the two equations

Take the origin at the launch point, x horizontal and y vertically upwards.

Direction	Acceleration	Velocity at time t	Position at time t
Horizontal	0	V cos α (constant)	x = (V cos α)t
Vertical	−g	V sin α − gt	y = (V sin α)t − ½gt²

Horizontal motion is uniform because nothing pushes sideways. Vertical motion has constant acceleration −g, so the constant-acceleration results from straight-line motion apply.

Two habits prevent most mistakes.

1. Treat the directions separately. Time t is the only link between them.
2. Fix a sign convention at the start: up is positive, so a launch below the horizontal has a negative vertical component.

Speed at time t is √((V cos α)² + (V sin α − gt)²). The direction of motion is tan θ = (vertical component)/(horizontal component), measured above or below the horizontal.

3. Time of flight, greatest height and range on level ground

You derive each result from the equations rather than quote it.

Time of flight. The particle returns to the launch level when y = 0:

(V sin α)t − ½gt² = 0
t((V sin α) − ½gt) = 0
t = 0 or t = 2V sin α / g

Greatest height. The vertical velocity is zero at the top, so V sin α − gt = 0 and t = V sin α / g, half the time of flight. Substituting gives y = (V sin α)²/(2g).

Range. Horizontal distance at the end of the flight is x = (V cos α)(2V sin α / g) = 2V² sin α cos α / g = V² sin 2α / g.

Two consequences follow from this. The range is greatest when sin 2α = 1, so α = 45°. Angles α and 90° − α give the same range, because sin 2α = sin(180° − 2α). The two paths have different heights and times of flight.

Worked example 1

A ball is projected from level ground with speed 24 m/s at 50° above the horizontal. Find the time of flight, the greatest height and the range.

Vertical component:   24 sin 50° = 18.385 m/s
Horizontal component: 24 cos 50° = 15.427 m/s
Time of flight: t = 2 × 18.385 / 9.8 = 3.752 s
Greatest height: y = 18.385² / (2 × 9.8) = 17.245 m
Range: x = 15.427 × 3.752 = 57.88 m

The time of flight is 3.75 s, the greatest height is 17.2 m and the range is 57.9 m (3 s.f.). A check is to find the range from V² sin 2α / g = 576 sin 100° / 9.8, which gives the same 57.9 m.

4. Launching from a height, or below the horizontal

If the ground is not at the launch level, do not use the level-ground results. Put the landing condition into the y equation and solve.

Worked example 2

A stone is thrown from the top of a cliff, 20 m above level ground, with speed 18 m/s at 25° above the horizontal. Find the time to reach the ground, the horizontal distance travelled, and the speed on landing.

Components: horizontal 18 cos 25° = 16.314, vertical 18 sin 25° = 7.607
Landing: y = −20, so  −20 = 7.607t − 4.9t²
4.9t² − 7.607t − 20 = 0
t = [7.607 + √(7.607² + 4 × 4.9 × 20)] / 9.8 = 2.941 s   (reject the negative root)
Horizontal distance: x = 16.314 × 2.941 = 47.97 m
Vertical velocity on landing: 7.607 − 9.8 × 2.941 = −21.21 m/s
Speed = √(16.314² + 21.21²) = 26.76 m/s

The stone lands after 2.94 s, 48.0 m from the foot of the cliff, at speed 26.8 m/s. The negative root is the time at which the extended path would have passed through ground level before launch, so it has no meaning here.

Worked example 3 (horizontal projection)

A stone is thrown horizontally at 16 m/s from the top of a cliff. It lands 40 m from the base of the cliff. Find the height of the cliff and the direction of motion on landing.

Here α = 0, so the initial vertical velocity is zero.

Horizontal: 40 = 16t, so t = 2.5 s
Vertical: drop = ½ × 9.8 × 2.5² = 30.625 m
Vertical velocity on landing: 9.8 × 2.5 = 24.5 m/s downwards
tan θ = 24.5 / 16, θ = 56.9° below the horizontal

The cliff is 30.6 m high and the stone lands moving at 56.9° below the horizontal, at speed √(16² + 24.5²) = 29.3 m/s.

5. Eliminating time: the equation of the trajectory

The path is a curve in x and y alone. To find it, eliminate t.

From x = (V cos α)t:        t = x / (V cos α)
Substitute into y = (V sin α)t − ½gt²:
y = (V sin α) x / (V cos α) − ½g x² / (V² cos² α)
y = x tan α − g x² / (2V² cos² α)

Using 1/cos² α = sec² α:

y = x tan α − g x² sec² α / (2V²)

The specification does not ask you to remember this, but you should be able to produce it. It is a downward-opening parabola. For problems where the angle is the unknown, use sec² α = 1 + tan² α to turn it into a quadratic in tan α.

Worked example 4 (finding the angle)

A ball is projected from O with speed 14 m/s. It passes through the point 10 m horizontally from O and 5 m above O. Find the possible angles of projection.

y = x tan α − g x² (1 + tan² α) / (2V²)
5 = 10T − 4.9 × 100 × (1 + T²) / 196        where T = tan α
5 = 10T − 2.5(1 + T²)
2.5T² − 10T + 7.5 = 0
T² − 4T + 3 = 0
(T − 1)(T − 3) = 0

So tan α = 1 or tan α = 3, giving α = 45° or α = 71.6° (1 d.p.). Two angles appear because the point can be reached on the way up or on the way down.

Worked example 5 (clearing a wall)

A ball is projected from level ground at 22 m/s at 30° above the horizontal. A vertical wall 4 m high stands 32 m from the launch point on the same level. Does the ball clear it?

Horizontal: 32 = 22 cos 30° × t, so t = 1.6796 s
Height at that time: y = 22 sin 30° × 1.6796 − 4.9 × 1.6796² = 18.476 − 13.823 = 4.65 m

The height is 4.65 m, which is greater than 4 m, so the ball clears the wall by 0.65 m. The vertical velocity at that moment is 11 − 9.8 × 1.6796 = −5.46 m/s, so the ball is on its way down. For a “does it clear” question, always compare heights at the horizontal distance of the obstacle, and say whether the ball is above or below it.

Common errors

• Using V instead of the components V cos α and V sin α in the equations.
• Using the level-ground range or time formulae when the landing point is above or below the launch point.
• Giving the speed at the greatest height as zero. Only the vertical component is zero there, so the speed is V cos α.
• Forgetting the sign: the landing condition from a cliff is y = −20, not 20.
• Rounding the time too early and losing accuracy in x. Keep full unrounded values and round at the end.
• Giving a time of flight without rejecting a negative root.
• Stating assumptions vaguely (“ignore friction”). Name the specific ones: air resistance ignored, particle model, g constant.

Where to go next

Practise with the projectile practice questions, condense with the revision notes and tick off the topic on the printable checklist. The hub lists the other units, and the diagnostics show which topics need attention first.

Official syllabus

OxfordAQA International AS and A-level Mathematics (9660), Version 5.2, published by OxfordAQA. Section M2.5: Projectiles.