# Projectile Motion

> Physics · CIE A-Level
> Source: https://www.owlsprep.com/study/cie-9702-u2-projectile-motion/

This module explains how to analyse 2D projectile motion under constant gravity, by splitting motion into independent horizontal and vertical components. You will learn to calculate key projectile properties and interpret real-world deviations from ideal motion.

**Prerequisites:** [Kinematic equations for constant acceleration](https://www.owlsprep.com/study/cie-9702-u2-constant-acceleration/); [Resolution of vectors into perpendicular components](https://www.owlsprep.com/study/cie-9702-u2-vectors/)

## Learning objectives

- Resolve projectile velocity into independent horizontal and vertical components
- Apply kinematic equations to each component of projectile motion
- Calculate time of flight, maximum height and range for common projectile scenarios
- Describe the effect of air resistance on ideal projectile motion

## Principle of Independent Motion

Projectile motion is 2-dimensional motion of an object given an initial velocity, that then moves only under the influence of constant gravitational force (ignoring air resistance for ideal motion). The core simplifying principle is that perpendicular components of motion are independent of each other.

**Principle of Independence of Motion** — Horizontal and vertical components of acceleration, velocity and displacement do not affect one another. They can be analysed separately and then combined to get the full motion.

*Example:* Horizontal acceleration is always 0 for ideal projectiles, this has no effect on vertical acceleration which always equals $g$, acting downwards.

**Worked example:** A projectile is launched with initial velocity $20 \text{ m s}^{-1}$ at an angle of $30^\text{o}$ to the horizontal. Resolve the initial velocity into horizontal and vertical components, taking upwards as positive.

1. Recall the resolution formula for a vector at angle $\theta$ to the horizontal:
2. $$u_x = u \cos\theta, \quad u_y = u \sin\theta$$
3. Substitute $u = 20 \, \text{m s}^{-1}$ and $\theta = 30^\circ$:
4. $$u_x = 20 \cos 30^\circ = 20 \times \frac{\sqrt{3}}{2} = 17.3 \, \text{m s}^{-1}$$
5. $$u_y = 20 \sin 30^\circ = 20 \times 0.5 = 10.0 \, \text{m s}^{-1}$$
6. Final answer: initial horizontal component is $17.3 \, \text{m s}^{-1}$ and initial vertical component is $10.0 \, \text{m s}^{-1}$ upwards.

> **Exam tip:** Always state your sign convention for vertical motion at the start of working. CIE examiners penalize avoidable sign errors.

## Calculations for Level Ground Launch

For a projectile launched from and landing at the same vertical height (level ground), we can use kinematic equations for each component to find time of flight, maximum height and range. Vertical motion is constant acceleration, horizontal motion is constant velocity.

**Worked example:** A ball is kicked from ground level with initial velocity $15 \, \text{m s}^{-1}$ at $40^\circ$ to the horizontal. Find (a) time of flight, (b) maximum height, (c) range. Take $g = 9.8 \, \text{m s}^{-2}$. Take upwards as positive.

1. First resolve the initial velocity:
2. $$u_x = 15 \cos 40^\circ = 11.49 \, \text{m s}^{-1}, \quad u_y = 15 \sin 40^\circ = 9.64 \, \text{m s}^{-1}$$
3. Part (a): Time of flight is when vertical displacement $s_y = 0$. Use $s = ut + \frac{1}{2} a t^2$:
4. $$0 = u_y t - \frac{1}{2} g t^2 = t\left(u_y - \frac{1}{2} g t\right)$$
5. Solutions are $t=0$ (launch) and $t = \frac{2u_y}{g}$:
6. $$t = \frac{2 \times 9.64}{9.8} = 1.97 \, \text{s}$$
7. Part (b): Maximum height occurs when vertical velocity $v_y = 0$. Use $v^2 = u^2 + 2as$:
8. $$0 = u_y^2 - 2gH \implies H = \frac{u_y^2}{2g} = \frac{(9.64)^2}{2 \times 9.8} = 4.74 \, \text{m}$$
9. Part (c): Range is horizontal displacement, with constant horizontal velocity:
10. $$R = u_x t = 11.49 \times 1.97 = 22.6 \, \text{m}$$

> **info**
>
> For ideal projectile motion on level ground, maximum range is always achieved at a projection angle of $45^\circ$.

## Projectiles Launched From a Height

A common CIE exam question involves projectiles launched from an initial height above the landing point, for example a ball thrown off a cliff. The same principle of independent components applies, but the final vertical displacement is non-zero.

**Worked example:** A stone is thrown horizontally at $8.0 \, \text{m s}^{-1}$ from the top of a 50 m tall cliff. Find how far from the base of the cliff it lands. Take $g = 9.8 \, \text{m s}^{-2}$.

1. Initial vertical velocity $u_y = 0$ because the stone is thrown horizontally. Take downwards as positive to simplify signs: initial vertical displacement $s_y = 50 \, \text{m}$, $a = g = 9.8 \, \text{m s}^{-2}$.
2. First calculate time of flight using vertical motion, with $s = ut + \frac{1}{2} a t^2$:
3. $$50 = (0)t + \frac{1}{2} \times 9.8 \times t^2$$
4. $$t^2 = \frac{100}{9.8} = 10.20 \implies t = 3.19 \, \text{s}$$
5. Horizontal velocity is constant, so calculate horizontal range:
6. $$R = u_x t = 8.0 \times 3.19 = 25.5 \, \text{m}$$

> **Exam tip:** If you take upwards as positive, remember the final vertical displacement when landing below the launch point is negative. Sign errors here are very common.

## Effect of Air Resistance

All calculations above assume ideal projectile motion with no air resistance. In real scenarios, air resistance creates a drag force opposite to the direction of motion, which alters the trajectory:

- Maximum height reached is lower than the ideal value
- Total range is shorter than the ideal prediction
- The trajectory is no longer symmetric: descent is steeper than ascent
- Horizontal velocity is not constant, it decreases over time due to deceleration from drag

**Check your understanding**

Test your understanding:

1. Which of the following statements correctly describes the effect of air resistance on an ideal projectile launched from level ground?

   - Maximum height is greater than the ideal value
   - Range is less than the ideal value
   - Horizontal velocity remains constant throughout the flight
   - The trajectory remains symmetric

   *Answer:* Range is less than the ideal value

   *Why:* Air resistance does negative work on the projectile, reducing its speed at every point. This leads to a shorter range and lower maximum height.

## Common pitfalls

- **Wrong:** Treating horizontal motion as accelerated, and applying constant acceleration kinematics to the horizontal component.
  - Why it fails: Only gravity acts on an ideal projectile, and gravity acts vertically only. There is no horizontal acceleration.
  - Correct: Always use constant velocity for horizontal motion: horizontal displacement = horizontal velocity × time of flight.
- **Wrong:** Using $a = +g$ when upwards is defined as the positive direction for vertical motion.
  - Why it fails: Gravity acts downwards, so it has a negative sign in this convention.
  - Correct: Write down your sign convention at the start of every projectile problem to avoid this error.
- **Wrong:** Using the range formula $R = \frac{u^2 \sin 2\theta}{g}$ for projectiles launched from a height.
  - Why it fails: This formula is only valid for projectiles that launch and land at the same vertical height.
  - Correct: Always start from first principles: solve for time using vertical motion, then calculate range from horizontal velocity × time.
- **Wrong:** Trying to calculate time of flight from horizontal motion instead of vertical motion.
  - Why it fails: Time of flight is determined by how long the projectile takes to move vertically to the landing point, not by horizontal motion.
  - Correct: Always solve for time using the vertical component of motion first, then use time to find horizontal range.
- **Wrong:** Assuming maximum range is always at 45°, even for projectiles launched from a height.
  - Why it fails: The 45° rule only applies to same-height launch and landing.
  - Correct: For launch from a height, the maximum range occurs at an angle less than 45°. Always confirm from first principles if asked.

## Cheatsheet

| Quantity | Formula (ideal motion, up positive) | Notes |
| --- | --- | --- |
| Initial horizontal component | $u_x = u \cos\theta$ | Constant for ideal motion |
| Initial vertical component | $u_y = u \sin\theta$ | Accelerated at $-g$ |
| Time of flight (level ground) | $t = \frac{2u_y}{g}$ | Same launch/landing height |
| Maximum height | $H = \frac{u_y^2}{2g}$ | $v_y = 0$ at max height |
| Range (level ground) | $R = \frac{u^2 \sin 2\theta}{g}$ | Max range at 45° |
| General case | $s_y = u_y t - \frac{1}{2} g t^2$ | Always works from first principles |

## What's next

Projectile motion is the foundational example of 2-dimensional motion in A-Level Physics, and the principle of resolving motion into independent perpendicular components is used repeatedly throughout the course. This technique extends to more complex topics including circular motion, collisions, and charged particle motion in uniform electric fields, where constant forces act in one dimension only. Projectile motion questions appear regularly in both multiple choice and structured sections of CIE 9702 papers, so mastering the method is key for consistent high scores.

- [Dynamics](https://www.owlsprep.com/study/cie-9702-u3-overview/)
- [Newton's laws of motion](https://www.owlsprep.com/study/cie-9702-u3-newton-s-laws-of-motion/)
- [Linear momentum](https://www.owlsprep.com/study/cie-9702-u3-linear-momentum/)

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