Do I need to memorise the formulae for range, time of flight, and the equation of the trajectory?

It is not essential, and it's much safer to derive them from the basic SUVAT equations each time. Understanding the derivation (e.g., setting $v_y=0$ for max height, $s_y=0$ for time of flight) is the key skill being tested. Memorised formulae only work for level ground and can lead to errors if misapplied.

What happens if the projectile lands at a different height from its launch point?

In this case, the final vertical displacement, $s_y$, is not zero. For example, if a particle is projected from the top of a 10m cliff and lands in the sea, its final vertical displacement is $s_y = -10 \, \text{m}$ (taking the launch point as the origin). You would use this value in the SUVAT equation $s_y = u_y t + \frac{1}{2}a_y t^2$ to find the time of flight.

Why is the path a parabola?

The equation of the trajectory, which we derived as $y = x \tan \theta - \frac{gx^2}{2U^2 \cos^2 \theta}$, is a quadratic function of $x$ (in the form $y = Ax - Bx^2$). The graph of a quadratic function is a parabola.

What if air resistance is not negligible?

If air resistance is considered, the problem becomes much more complex. Air resistance is a type of drag force that opposes motion. It is often modelled as being proportional to velocity or the square of velocity. This means the acceleration is no longer constant, and the SUVAT equations do not apply. You would need to use calculus (forming and solving differential equations) to analyse the motion. This is covered in more advanced topics within Mechanics.

Where can I learn Motion of a projectile for Cambridge 9231?

This free lesson on MarkScheme covers syllabus point 3.1 (Motion of a projectile) for Cambridge A-Level Further Mathematics. It includes visual revision, key points, and links to real past-paper questions with mark-scheme marking at markscheme.app/courses/9231/3-1-motion-of-a-projectile.

Is Motion of a projectile (3.1) on 9231 free to study?

Yes. The Further Mathematics 9231 course on MarkScheme is 100% free — including premium visual lessons, exam tips, and past-paper marking. No paywall for Motion of a projectile.

Motion of a projectile 9231 — Free A-Level…

Motion of a projectile

Welcome to the study of projectile motion in Further Mathematics. We will build on your existing knowledge of kinematics and SUVAT equations from AS-level Mechanics. The key difference at this level is the introduction of more complex scenarios, most notably projectiles moving relative to an inclined plane. The fundamental principle remains the same: we can analyse the motion by considering the horizontal and vertical components independently, assuming negligible air resistance.

BY THE END, YOU CAN…

1Model the motion of a projectile as a particle moving with constant acceleration.
2Resolve initial velocity into horizontal and vertical components.
3Calculate the time of flight, range, and greatest height of a projectile.
4Solve problems involving projectiles on an inclined plane by resolving gravity.

In simple terms

A friendly intro before the formal notes — no formulas yet.

The Parabolic Path

We analyse the curved path of a projectile by splitting its motion into two independent parts. The horizontal motion is steady, while the vertical motion is affected by gravity.

THINK OF IT LIKE…

Imagine you're on a moving travelator at an airport and you throw a ball straight up. To you, the ball just goes up and comes back down. But to a stationary observer, the ball traces a curved path, moving forward at the travelator's constant speed while also accelerating up and down due to your throw and gravity. Projectile motion works the same way: we analyse the constant horizontal 'travelator' motion and the accelerated vertical 'ball toss' motion separately.

1
Resolve the initial velocity into horizontal ( $u_x$ ) and vertical ( $u_y$ ) components using trigonometry.
2
Analyse the vertical motion using SUVAT equations with acceleration $a_y = -g$ to find key timings, like time to maximum height or total time of flight.
3
Analyse the horizontal motion using the simple formula distance = speed × time, as horizontal acceleration $a_x = 0$ .
4
Combine the results from the horizontal and vertical analyses to find quantities like the range or the position at a given time.

Explore the concept

Use the live diagram and synced steps — play it or tap a step card to walk through.

Key formulas

Tap any symbol to reveal exactly what it means and its units.

$u_x = U \cos \theta$

Tap a symbol — great for exam definitions

$\alpha$

Tap a symbol — great for exam definitions

$x = (U \cos \theta) t \implies t = \frac{x}{U \cos \theta}$

Tap a symbol — great for exam definitions

Full topic notes

Formal explanation with the rigour you need for the exam.

1. Decomposing Motion: The Foundation of Projectiles

Any object thrown or projected into the air, subject only to the force of gravity, is a projectile. Its path is a parabola. To analyse this, we resolve the motion into two perpendicular components. Conventionally, we use horizontal ( $x$ ) and vertical ( $y$ ) axes. The initial velocity, $U$ , at an angle $\theta$ to the horizontal is split into components.

Initial horizontal velocity: $u_x = U \cos \theta$ \ Initial vertical velocity: $u_y = U \sin \theta$

Horizontal Motion: No forces act horizontally (as air resistance is ignored), so acceleration is zero ( $a_x = 0$ ). The horizontal velocity is constant. The only equation needed is: $s_x = u_x t$ .
Vertical Motion: Gravity acts downwards, so acceleration is constant ( $a_y = -g$ , taking upwards as positive). The standard SUVAT equations apply to the vertical component of motion.
Time ( $t$ ) is the scalar quantity that links the horizontal and vertical analyses. The time taken to reach a certain horizontal position is the same as the time taken to reach the corresponding vertical position.

2. Projectiles on an Inclined Plane

A common extension in Further Mathematics is to analyse a projectile launched from, and landing on, a sloped surface. For these problems, it is much more convenient to rotate our coordinate system. We define our axes to be parallel and perpendicular to the inclined plane itself. This simplifies the displacement calculations, but it means we must resolve the acceleration due to gravity, $g$ , into components in our new axis system.

For a plane inclined at angle $\alpha$ to the horizontal: Acceleration parallel to the plane (downslope): $a_\parallel = g \sin \alpha$ Acceleration perpendicular to the plane (into the plane): $a_\perp = g \cos \alpha$

EXAM TIP

Always draw a clear diagram for inclined plane problems. Label the angle of the plane, $\alpha$ , and the angle of projection relative to the plane. Be extremely careful when resolving both the initial velocity and the acceleration due to gravity. A common mistake is to mix up $\sin$ and $\cos$ for the acceleration components.

3. Equation of the Trajectory

Sometimes, you may be asked to find the Cartesian equation of the projectile's path. This equation relates the vertical displacement, $y$ , to the horizontal displacement, $x$ , without involving time, $t$ . To derive this, we express $t$ in terms of $x$ from the horizontal motion equation and substitute it into the vertical motion equation.

Horizontal motion: $x = (U \cos \theta) t \implies t = \frac{x}{U \cos \theta}$
Vertical motion: $y = (U \sin \theta) t - \frac{1}{2}gt^2$
Substitute for $t$ : $y = (U \sin \theta) \left(\frac{x}{U \cos \theta}\right) - \frac{1}{2}g\left(\frac{x}{U \cos \theta}\right)^2$ This simplifies to the parabolic form: $y = x \tan \theta - \frac{gx^2}{2U^2 \cos^2 \theta}$

The equation of the trajectory is a quadratic in $x$ , which confirms the path is a parabola.
This form is useful for problems where you are given a point $(x, y)$ that the projectile passes through and need to find the angle of projection $\theta$ or initial speed $U$ .
Remember the identity $\sec^2 \theta = 1 + \tan^2 \theta$ . Substituting this can transform the equation into a quadratic in $\tan \theta$ , which is often easier to solve.

Worked examples

See the formulas applied — reveal one step at a time, like the exam.

EXAMPLE 1

A golf ball is struck from a point on level horizontal ground with a speed of $42 \, \text{m s}^{-1}$ at an angle of $30^\circ$ above the horizontal. Find: (a) the greatest height reached by the ball. (b) the time of flight. (c) the range of the ball on the horizontal ground. (Use $g = 9.8 \, \text{m s}^{-2}$ )

1
First, resolve the initial velocity into components. Let upwards and rightwards be positive. $u_x = 42 \cos 30^\circ = 42 \times \frac{\sqrt{3}}{2} = 21\sqrt{3} \, \text{m s}^{-1}$ $u_y = 42 \sin 30^\circ = 42 \times \frac{1}{2} = 21 \, \text{m s}^{-1}$

EXAMPLE 2

A particle is projected from a point $O$ on a plane inclined at $20^\circ$ to the horizontal. The initial velocity is $30 \, \text{m s}^{-1}$ at an angle of $45^\circ$ to the plane. The particle moves up the plane. Find: (a) the time of flight. (b) the range of the particle along the plane. (Use $g = 9.8 \, \text{m s}^{-2}$ )

1
Let's set up a coordinate system with the x-axis parallel to the plane (upwards) and the y-axis perpendicular to the plane (outwards).

How it all connects

The big idea sits in the middle — tap a linked idea to explore the link.

MAIN IDEAMotion of a projectile

Tap a linked idea to see how it connects back to the main topic — that connection is what examiners reward.

Glossary

Try to recall each definition before you reveal it.

Quick check

Answer in your head first — then tap to check. No pressure.

Revision flashcards

Flip the card. Test yourself before the exam.

QUESTION · 1 / 11

What is the key assumption made when modelling projectile motion?

TAP OR SPACE TO FLIP

ANSWER

Air resistance is negligible, and the acceleration due to gravity, $g$ , is constant.

TAP OR SPACE TO FLIP BACK

1 / 11 · ← → space

Key takeaways

Review these before you close the topic — retrieval beats re-reading.

✓
Horizontal Motion: No forces act horizontally (as air resistance is ignored), so acceleration is zero ( $a_x = 0$ ). The horizontal velocity is constant. The only equation needed is: $s_x = u_x t$ .
✓
Vertical Motion: Gravity acts downwards, so acceleration is constant ( $a_y = -g$ , taking upwards as positive). The standard SUVAT equations apply to the vertical component of motion.
✓
Time ( $t$ ) is the scalar quantity that links the horizontal and vertical analyses. The time taken to reach a certain horizontal position is the same as the time taken to reach the corresponding vertical position.

Practice — then mark it

The whole point: a real Cambridge question, marked mark-by-mark.

Real past paper[0]

Test your understanding with past paper questions

Do it on paper → mark itMARKED MARK-BY-MARK · B1 / M1 / A1 · OFFICIAL SCHEME

Extra simulations & links

PhET, GeoGebra and other curated tools — open in a new tab.

Frequently asked

✓

Checkpoint

One marked question is worth ten re-reads — close the loop before you move on.

Checkpoint

Reading it isn’t knowing it — prove it.

Before you move on: do Test your understanding with past paper questions on paper, snap a photo, and get examiner-style feedback on exactly where you win and lose marks.

Attempt & get marked →Takes about a minute · you’ll land back on this lesson

← PREVIOUS2.6 Differential equations NEXT →3.2 Equilibrium of a rigid body

Motion of a projectile

BY THE END, YOU CAN…

1Model the motion of a projectile as a particle moving with constant acceleration.
2Resolve initial velocity into horizontal and vertical components.
3Calculate the time of flight, range, and greatest height of a projectile.
4Solve problems involving projectiles on an inclined plane by resolving gravity.

In simple terms

A friendly intro before the formal notes — no formulas yet.

The Parabolic Path

We analyse the curved path of a projectile by splitting its motion into two independent parts. The horizontal motion is steady, while the vertical motion is affected by gravity.

THINK OF IT LIKE…

1
Resolve the initial velocity into horizontal ( $u_x$ ) and vertical ( $u_y$ ) components using trigonometry.
2
Analyse the vertical motion using SUVAT equations with acceleration $a_y = -g$ to find key timings, like time to maximum height or total time of flight.
3
Analyse the horizontal motion using the simple formula distance = speed × time, as horizontal acceleration $a_x = 0$ .
4
Combine the results from the horizontal and vertical analyses to find quantities like the range or the position at a given time.

Explore the concept

Use the live diagram and synced steps — play it or tap a step card to walk through.

Key formulas

Tap any symbol to reveal exactly what it means and its units.

$u_x = U \cos \theta$

Tap a symbol — great for exam definitions

$\alpha$

Tap a symbol — great for exam definitions

$x = (U \cos \theta) t \implies t = \frac{x}{U \cos \theta}$

Tap a symbol — great for exam definitions

Full topic notes

Formal explanation with the rigour you need for the exam.

1. Decomposing Motion: The Foundation of Projectiles

Initial horizontal velocity: $u_x = U \cos \theta$ \ Initial vertical velocity: $u_y = U \sin \theta$

Horizontal Motion: No forces act horizontally (as air resistance is ignored), so acceleration is zero ( $a_x = 0$ ). The horizontal velocity is constant. The only equation needed is: $s_x = u_x t$ .
Vertical Motion: Gravity acts downwards, so acceleration is constant ( $a_y = -g$ , taking upwards as positive). The standard SUVAT equations apply to the vertical component of motion.
Time ( $t$ ) is the scalar quantity that links the horizontal and vertical analyses. The time taken to reach a certain horizontal position is the same as the time taken to reach the corresponding vertical position.

2. Projectiles on an Inclined Plane

EXAM TIP

3. Equation of the Trajectory

Horizontal motion: $x = (U \cos \theta) t \implies t = \frac{x}{U \cos \theta}$
Vertical motion: $y = (U \sin \theta) t - \frac{1}{2}gt^2$
Substitute for $t$ : $y = (U \sin \theta) \left(\frac{x}{U \cos \theta}\right) - \frac{1}{2}g\left(\frac{x}{U \cos \theta}\right)^2$ This simplifies to the parabolic form: $y = x \tan \theta - \frac{gx^2}{2U^2 \cos^2 \theta}$

The equation of the trajectory is a quadratic in $x$ , which confirms the path is a parabola.
This form is useful for problems where you are given a point $(x, y)$ that the projectile passes through and need to find the angle of projection $\theta$ or initial speed $U$ .
Remember the identity $\sec^2 \theta = 1 + \tan^2 \theta$ . Substituting this can transform the equation into a quadratic in $\tan \theta$ , which is often easier to solve.

Worked examples

See the formulas applied — reveal one step at a time, like the exam.

EXAMPLE 1

1
First, resolve the initial velocity into components. Let upwards and rightwards be positive. $u_x = 42 \cos 30^\circ = 42 \times \frac{\sqrt{3}}{2} = 21\sqrt{3} \, \text{m s}^{-1}$ $u_y = 42 \sin 30^\circ = 42 \times \frac{1}{2} = 21 \, \text{m s}^{-1}$

EXAMPLE 2

1
Let's set up a coordinate system with the x-axis parallel to the plane (upwards) and the y-axis perpendicular to the plane (outwards).

How it all connects

The big idea sits in the middle — tap a linked idea to explore the link.

MAIN IDEAMotion of a projectile

Tap a linked idea to see how it connects back to the main topic — that connection is what examiners reward.

Glossary

Try to recall each definition before you reveal it.

Quick check

Answer in your head first — then tap to check. No pressure.

Revision flashcards

Flip the card. Test yourself before the exam.

QUESTION · 1 / 11

What is the key assumption made when modelling projectile motion?

TAP OR SPACE TO FLIP

ANSWER

Air resistance is negligible, and the acceleration due to gravity, $g$ , is constant.

TAP OR SPACE TO FLIP BACK

1 / 11 · ← → space

Key takeaways

Review these before you close the topic — retrieval beats re-reading.

✓
Horizontal Motion: No forces act horizontally (as air resistance is ignored), so acceleration is zero ( $a_x = 0$ ). The horizontal velocity is constant. The only equation needed is: $s_x = u_x t$ .
✓
Vertical Motion: Gravity acts downwards, so acceleration is constant ( $a_y = -g$ , taking upwards as positive). The standard SUVAT equations apply to the vertical component of motion.
✓
Time ( $t$ ) is the scalar quantity that links the horizontal and vertical analyses. The time taken to reach a certain horizontal position is the same as the time taken to reach the corresponding vertical position.

Practice — then mark it

The whole point: a real Cambridge question, marked mark-by-mark.

Real past paper[0]

Test your understanding with past paper questions

Do it on paper → mark itMARKED MARK-BY-MARK · B1 / M1 / A1 · OFFICIAL SCHEME

Extra simulations & links

PhET, GeoGebra and other curated tools — open in a new tab.

Frequently asked

✓

Checkpoint

One marked question is worth ten re-reads — close the loop before you move on.

Checkpoint

Reading it isn’t knowing it — prove it.

Before you move on: do Test your understanding with past paper questions on paper, snap a photo, and get examiner-style feedback on exactly where you win and lose marks.

Attempt & get marked →Takes about a minute · you’ll land back on this lesson

← PREVIOUS2.6 Differential equations NEXT →3.2 Equilibrium of a rigid body

Revise Motion of a projectile for Cambridge 9231 (3.1)

In simple terms

Explore the concept

Key formulas

Full topic notes

1. Decomposing Motion: The Foundation of Projectiles

2. Projectiles on an Inclined Plane

3. Equation of the Trajectory

Worked examples

How it all connects

Glossary

Quick check

Revision flashcards

Key takeaways

Practice — then mark it

Test your understanding with past paper questions

Extra simulations & links

Frequently asked

Checkpoint

Reading it isn’t knowing it — prove it.

In simple terms

Explore the concept

Key formulas

Full topic notes

1. Decomposing Motion: The Foundation of Projectiles

2. Projectiles on an Inclined Plane

3. Equation of the Trajectory

Worked examples

How it all connects

Glossary

Quick check

Revision flashcards

Key takeaways

Practice — then mark it

Test your understanding with past paper questions

Extra simulations & links

Frequently asked

Checkpoint

Reading it isn’t knowing it — prove it.

Discuss Motion of a projectile