Motion of a projectile

Further Mathematics· 9231 Unit 3: Further Mechanics, Section 2: Projectiles· 25 min read

1. Core Kinematic Equations for Projectile Motion★★★☆☆⏱ 10 min

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Projectile motion assumes no air resistance, so the only force acting on the particle is its weight, directed vertically downwards. This means horizontal acceleration is exactly 0, and vertical acceleration is constant at $- g$ if upward is defined as the positive direction.

📘 Definition

Oblique Projection

$u_{x} = u cos θ, u_{y} = u sin θ$

A particle launched with initial speed $u$ at angle $θ$ above the horizontal, with initial velocity resolved into independent horizontal and vertical components.

🔬 Derivation

Goal:

Derive displacement components for projectile motion

Starting from:

Initial velocity components $u_{x} = u cos θ$ , $u_{y} = u sin θ$ , $a_{x} = 0$ , $a_{y} = - g$

1
Apply constant acceleration formula for horizontal displacement, no acceleration:
2
$x = u_x t = ut \cos\theta$
3
Apply constant acceleration formula for vertical displacement, acceleration $- g$ :
4
$y = u_y t - \frac{1}{2} g t^2 = ut \sin\theta - \frac{1}{2} g t^2$

Result:

Horizontal and vertical motion are fully independent of each other, and can be solved separately.

📐 Worked Example

A projectile is launched at 20 m/s at 30° above horizontal. Find its horizontal and vertical displacement after 1.5 seconds.

1
Resolve initial velocity components:
2
$u_x = 20 \cos 30^\circ = 10\sqrt{3} \approx 17.32 \text{ m s}^{-1}, u_y = 20 \sin 30^\circ = 10 \text{ m s}^{-1}$
3
Calculate horizontal displacement (no acceleration):
4
$x = 17.32 \times 1.5 = 26.0 \text{ m (3 s.f.)}$
5
Calculate vertical displacement:
6
$y = 10 \times 1.5 - 0.5 \times 9.8 \times (1.5)^2 = 15 - 11.025 = 3.98 \text{ m (3 s.f.)}$

2. Trajectory, Maximum Height and Horizontal Range★★★☆☆⏱ 8 min

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Exam tip:

CIE almost always asks you to derive the trajectory equation before using it for further calculations.

3. Projectile Motion on Inclined Planes★★★★☆⏱ 12 min

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For a projectile landing on an upward sloped plane at angle $α$ to the horizontal, the landing point satisfies the relation $y = x tan α$ , as the vertical height of the incline at horizontal position $x$ is $x tan α$ . Substitute this into the standard trajectory equation to solve for the point of intersection.

📐 Worked Example

A projectile is launched at 15 m/s at 45° above horizontal onto an incline of 20° above the horizontal. Find the range measured along the incline.

1
Write the trajectory equation for this projectile:
2
$y = x \tan 45^\circ - \frac{9.8 x^2}{2 \times 15^2 \cos^2 45^\circ} = x - \frac{9.8 x^2}{225}$
3
Substitute $y = x tan 2 0^{\circ} \approx 0.364 x$ :
4
$0.364x = x - 0.0436 x^2 \implies 0.636 x = 0.0436 x^2$
5
Cancel non-zero $x$ term to find horizontal landing position:
6
$x = 0.636 / 0.0436 \approx 14.59 \text{ m}$
7
Calculate range along the incline:
8
$R_{\text{incline}} = \frac{x}{\cos 20^\circ} \approx 15.5 \text{ m (3 s.f.)}$

4. Optimization of Projectile Parameters★★★★☆⏱ 10 min

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✓ Quick check

Test your understanding of optimization rules:

What is the optimal launch angle for maximum range on a horizontal plane with equal launch and landing height?
- 30°
- 45°
- 60°
- 90°
Reveal answer
45° —
The $sin 2 θ$ term in the horizontal range formula peaks at $2 θ = 9 0^{\circ}$ , so $θ = 4 5^{\circ}$ .

5. Common Pitfalls

Wrong move:

Using $g = + 9.8$ when defining upward as the positive vertical direction

Why:

Sign convention mismatch leads to negative height and time values that are physically impossible

Correct move:

Explicitly state your coordinate axes at the start of every problem, and set vertical acceleration to $- g$ if upward is positive

Wrong move:

Using the standard horizontal range formula directly for projectiles landing on inclined planes

Why:

The standard $R = u^{2} sin 2 θ / g$ result only applies when launch and landing heights are exactly equal

Correct move:

Substitute the incline relation $y = x tan α$ into the trajectory equation to solve for the landing position from first principles

Wrong move:

Resolving gravitational acceleration into horizontal and vertical components

Why:

Gravitational force acts purely vertically, there is no horizontal component of acceleration for unpowered projectiles with no air resistance

Correct move:

Set horizontal acceleration to 0 for all no-resistance projectile problems

Wrong move:

Assuming maximum range is always at 45° for projectiles launched from a height different to their landing height

Why:

The 45° optimal angle result is a special case that only holds when launch and landing heights are identical

Correct move:

Differentiate the range expression with respect to $θ$ and set the derivative to 0 to find the optimal angle for non-standard boundary conditions

Wrong move:

Including mass terms in projectile kinematic calculations

Why:

Gravitational acceleration is independent of mass, so the motion of a projectile under no air resistance does not depend on its mass at all

Correct move:

Cancel all mass terms early in your working to simplify calculations

6. Quick Reference Cheatsheet

Quantity	Horizontal Plane (equal launch/landing height)	Upward Inclined Plane (angle $α$ )
Time of flight	$\frac{2 u s i n θ}{g}$	$\frac{2 u s i n ( θ - α )}{g c o s α}$
Maximum height above launch	$\frac{u ^{2} s i n ^{2} θ}{2 g}$	$\frac{u ^{2} s i n ^{2} ( θ - α )}{2 g c o s ^{2} α}$
Total range	$\frac{u ^{2} s i n 2 θ}{g}$	$\frac{2 u ^{2} s i n ( θ - α ) c o s θ}{g c o s ^{2} α}$
Optimal launch angle for max range	$4 5^{\circ}$	$4 5^{\circ} + \frac{α}{2}$

7. Frequently Asked

Do I get marks for quoting standard projectile formulas directly in the exam?

CIE examiners award full marks for standard results only if the question says 'write down'. For all other questions, you must derive results from first principles using $s = u t + \frac{1}{2} a t^{2}$ to earn full method marks.

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2024 · Paper 4
Inclined plane projectile range
2023 · Paper 3
Projectile trajectory derivation
2022 · Paper 4
Optimal launch angle calculation

What's Next

Mastering projectile motion is a critical foundation for all further kinematics topics in CIE A-Level Further Maths. You will combine the vector resolution and constant acceleration techniques you learned here with work-energy principles to solve complex 15+ mark extended response questions that regularly appear on Paper 3 and Paper 4. Before moving on, ensure you can derive all standard projectile results from first principles without referencing the cheatsheet, as CIE examiners explicitly award marks for full derivation steps. This knowledge will extend directly to motion with resistive forces, circular motion, and relative kinematics.