# Motion of a projectile

> Further Mathematics · CIE A-Level
> Source: https://www.owlsprep.com/study/cie-9231-u3-motion-of-a-projectile/

This module covers 2D projectile motion under uniform gravity, including core kinematic derivations, trajectory equations, range on horizontal and inclined planes, and optimization of launch parameters.

**Prerequisites:** [1D kinematics under constant acceleration](https://www.owlsprep.com/study/cie-9231-u3-1d-constant-acceleration/); [2D vector resolution of forces and velocities](https://www.owlsprep.com/study/cie-9702-m1-vector-resolution/)

## Learning objectives

- Derive core projectile motion equations from first principles under uniform gravitational acceleration
- Calculate trajectory, maximum height, time of flight and range for projectiles on horizontal planes
- Solve projectile motion problems on upward and downward inclined planes
- Find optimal launch angles to maximize range for different boundary conditions

## Core Kinematic Equations for Projectile Motion

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.

**Oblique Projection** — A particle launched with initial speed $u$ at angle $\theta$ above the horizontal, with initial velocity resolved into independent horizontal and vertical components.

*Notation:* $u_x = u\cos\theta, u_y = u\sin\theta$

**Derivation:** Derive displacement components for projectile motion

*Starting from:* Initial velocity components $u_x = u\cos\theta$, $u_y = u\sin\theta$, $a_x = 0$, $a_y = -g$

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. $$y = u_y t - \frac{1}{2} g t^2 = ut \sin\theta - \frac{1}{2} g t^2$$

*Conclusion:* 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. $$y = 10 \times 1.5 - 0.5 \times 9.8 \times (1.5)^2 = 15 - 11.025 = 3.98 \text{ m (3 s.f.)}$$

*Calculator:* forbidden

## Trajectory, Maximum Height and Horizontal Range

> **Exam tip:** CIE almost always asks you to derive the trajectory equation before using it for further calculations.

*Calculator:* allowed

## Projectile Motion on Inclined Planes

For a projectile landing on an upward sloped plane at angle $\alpha$ to the horizontal, the landing point satisfies the relation $y = x \tan \alpha$, as the vertical height of the incline at horizontal position $x$ is $x \tan \alpha$. 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 $y = x \tan 20^\circ \approx 0.364 x$:
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. $$R_{\text{incline}} = \frac{x}{\cos 20^\circ} \approx 15.5 \text{ m (3 s.f.)}$$

*Calculator:* allowed

## Optimization of Projectile Parameters

> **Incline Optimal Angle Rule**
>
> For maximum range up an incline of angle $\alpha$, the optimal launch angle is always halfway between the vertical and the incline surface, i.e. $45^\circ + \frac{\alpha}{2}$.

**Check your understanding**

Test your understanding of optimization rules:

1. What is the optimal launch angle for maximum range on a horizontal plane with equal launch and landing height?

   - 30°
   - 45°
   - 60°
   - 90°

   *Why:* The $\sin 2\theta$ term in the horizontal range formula peaks at $2\theta = 90^\circ$, so $\theta = 45^\circ$.

*Calculator:* forbidden

## Common pitfalls

- **Wrong:** Using $g = +9.8$ when defining upward as the positive vertical direction
  - Why it fails: Sign convention mismatch leads to negative height and time values that are physically impossible
  - Correct: Explicitly state your coordinate axes at the start of every problem, and set vertical acceleration to $-g$ if upward is positive
- **Wrong:** Using the standard horizontal range formula directly for projectiles landing on inclined planes
  - Why it fails: The standard $R = u^2 \sin 2\theta / g$ result only applies when launch and landing heights are exactly equal
  - Correct: Substitute the incline relation $y = x \tan \alpha$ into the trajectory equation to solve for the landing position from first principles
- **Wrong:** Resolving gravitational acceleration into horizontal and vertical components
  - Why it fails: Gravitational force acts purely vertically, there is no horizontal component of acceleration for unpowered projectiles with no air resistance
  - Correct: Set horizontal acceleration to 0 for all no-resistance projectile problems
- **Wrong:** Assuming maximum range is always at 45° for projectiles launched from a height different to their landing height
  - Why it fails: The 45° optimal angle result is a special case that only holds when launch and landing heights are identical
  - Correct: Differentiate the range expression with respect to $\theta$ and set the derivative to 0 to find the optimal angle for non-standard boundary conditions
- **Wrong:** Including mass terms in projectile kinematic calculations
  - Why it fails: 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: Cancel all mass terms early in your working to simplify calculations

## Cheatsheet

| Quantity | Horizontal Plane (equal launch/landing height) | Upward Inclined Plane (angle $\alpha$) |
| --- | --- | --- |
| Time of flight | $\frac{2u\sin\theta}{g}$ | $\frac{2u\sin(\theta-\alpha)}{g\cos\alpha}$ |
| Maximum height above launch | $\frac{u^2 \sin^2 \theta}{2g}$ | $\frac{u^2 \sin^2(\theta-\alpha)}{2g \cos^2 \alpha}$ |
| Total range | $\frac{u^2 \sin 2\theta}{g}$ | $\frac{2u^2 \sin(\theta-\alpha) \cos \theta}{g \cos^2 \alpha}$ |
| Optimal launch angle for max range | $45^\circ$ | $45^\circ + \frac{\alpha}{2}$ |

## 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.

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