# Free fall

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

This sub-topic covers motion of objects under constant gravitational acceleration, including dropped objects and vertical projection. You will learn to solve calculation problems and interpret free fall motion for exam questions.

**Prerequisites:** [Constant acceleration kinematic equations](https://www.owlsprep.com/study/cie-9702-u2-constant-acceleration/)

## Learning objectives

- Define free fall and state its core assumptions
- Apply kinematic equations to solve free fall problems for dropped and projected objects
- Calculate maximum height and time of flight for vertically projected objects
- Describe experiments to determine acceleration due to gravity $g$ via free fall

## Core definition and properties of free fall

**Free fall** — Free fall describes any motion where the only force acting on an object is gravity. Air resistance is assumed negligible in all CIE problems unless explicitly stated otherwise.

*Example:* A ball dropped from rest off a cliff is in free fall until it impacts the ground.

Near Earth's surface, all objects in free fall experience the same constant acceleration, $g$, regardless of mass. For all CIE A-Level Physics exams, you should use $g = 9.81 \text{ m s}^{-2}$ unless the question specifies $10 \text{ m s}^{-2}$.

> **tip**
>
> A common misconception is that heavier objects fall faster. In free fall, all masses accelerate at exactly the same rate!

**Check your understanding**

Check your understanding

1. A skydiver reaches constant terminal velocity after jumping out of a plane. Is she in free fall at terminal velocity?

   - Yes, because she is falling downwards
   - No, because air resistance balances gravity, so gravity is not the only force

   *Answer:* No, because air resistance balances gravity, so gravity is not the only force

   *Why:* Correct! Free fall only occurs when gravity is the only force acting on the object. At terminal velocity, air resistance cancels gravity, so this is not free fall.

## Problem solving: Objects dropped from rest

For dropped objects, we use the standard constant acceleration kinematic equations, substituting $a = g$ and following a consistent sign convention (we will use downwards as positive for simplicity here).

**Worked example:** A ball is dropped from rest from the top of a 45 m tall building. Calculate the time taken to hit the ground and final impact velocity, using $g = 9.81 \text{ m s}^{-2}$.

1. Step 1: List all known values
2. $$u = 0 \text{m s}^{-1}, \\ s = 45 \text{m}, \\ a = g = 9.81 \text{m s}^{-2}, \\ t = ?, \\ v = ?$$
3. Step 2: Use $s = ut + \frac{1}{2} a t^2$. Since $u=0$, this simplifies to:
4. $$s = \frac{1}{2} g t^2 \implies t = \sqrt{\frac{2s}{g}}$$
5. Step 3: Substitute values to find t:
6. $$t = \sqrt{\frac{2 \times 45}{9.81}} = \sqrt{9.17} \approx 3.03 \text{ s}$$
7. Step 4: Use $v = u + at$ to find final velocity:
8. $$v = 0 + (9.81)(3.03) \approx 29.7 \text{ m s}^{-1}$$

> **tip**
>
> Always list your known values before selecting an equation. This avoids wasting time on the wrong equation for your given variables.

## Vertical projection: Objects thrown upwards

When an object is thrown straight upwards, it still accelerates downwards at $g = 9.81 \text{ m s}^{-2}$. At maximum height, the object's vertical velocity equals zero, before it accelerates back down to the ground.

**Worked example:** A ball is thrown vertically upwards from ground level with an initial speed of $15 \text{ m s}^{-1}$. Calculate (a) the maximum height reached, and (b) the total time taken to return to ground level.

1. Step 1: Take upwards as positive, list known values:
2. $$u = 15 \text{m s}^{-1}, \\ a = -g = -9.81 \text{m s}^{-2}, \\ v = 0 \text{m s}^{-1} \text{ (at max height)}$$
3. Step 2: For maximum height, use $v^2 = u^2 + 2as$:
4. $$0 = (15)^2 + 2(-9.81)s \implies 19.62s = 225 \implies s \approx 11.5 \text{ m}$$
5. Step 3: For total time, displacement when returning to ground is $s=0$. Use $s = ut + \frac{1}{2} a t^2$:
6. $$0 = 15t + \frac{1}{2}(-9.81)t^2 = t(15 - 4.905t)$$
7. Solutions are $t=0$ (initial throw) and:
8. $$t = \frac{15}{4.905} \approx 3.06 \text{ s}$$

Notice that when starting and ending at the same height, the time to reach maximum height is half the total flight time, and the speed on return equals the initial projection speed (in the opposite direction).

## Experimental measurement of $g$

CIE regularly asks for descriptions of free fall experiments to measure acceleration due to gravity. The most common method uses an electromagnet, timer and light gate:

1. A steel ball is held by an electromagnet at a measured height $h$ above a light gate.
2. When the electromagnet is switched off, the ball starts falling and the timer starts.
3. The timer stops when the ball passes through the light gate, giving time of fall $t$.
4. Plot a graph of $h$ against $t^2$. From $h = \frac{1}{2} g t^2$, gradient = $\frac{g}{2}$, so $g = 2 \times \text{gradient}$.

> **info**
>
> Other valid methods include ticker tape for a falling mass, or strobe photography. All use graphing to find $g$ from the gradient of a linear plot.

## Common pitfalls

- **Wrong:** Forgetting to make $g$ negative when upwards is taken as positive
  - Why it fails: Acceleration always acts downwards, so an incorrect sign leads to wrong values for height and time
  - Correct: Always confirm your sign convention before substituting acceleration into kinematic equations
- **Wrong:** Assuming displacement is zero at maximum height for upward projection
  - Why it fails: Only velocity is zero at maximum height, displacement is non-zero and equal to the maximum height
  - Correct: Remember: $v = 0$ at maximum height, $s \neq 0$ unless you start and end at the same point
- **Wrong:** Using $g = 10 \text{ m s}^{-2}$ when the question does not specify this value
  - Why it fails: CIE examiners penalize answers that use the wrong value of $g$ when 9.81 is expected
  - Correct: Always use $g = 9.81 \text{ m s}^{-2}$ unless the question explicitly tells you to use 10
- **Wrong:** Stopping calculation at maximum height for a ball thrown upwards off a cliff
  - Why it fails: You need to add the time taken to fall from maximum height to the cliff base to get total flight time
  - Correct: Always check the start and end positions to confirm you calculate the full displacement for the motion
- **Wrong:** Ignoring air resistance when the question says it is not negligible
  - Why it fails: Free fall assumptions only apply when air resistance is negligible, so acceleration is not constant if air resistance is significant
  - Correct: Always read the question carefully to check for mentions of air resistance before applying constant acceleration equations

## Cheatsheet

| Scenario | Key Rule/Equation | Notes |
| --- | --- | --- |
| Dropped from rest | $s = \frac{1}{2} g t^2$, $v = gt$ | $u=0$, downwards positive |
| Maximum height (upward projection) | $v^2 = u^2 - 2gs$, $v=0$ | Upwards positive, $a=-g$ |
| Total flight time (back to start) | $t = \frac{2u}{g}$ | Start/end at same displacement |
| Experimental $g$ from $h$ vs $t^2$ | $g = 2 \times \text{graph gradient}$ | Derived from $h = \frac{1}{2} g t^2$ |
| Acceleration at maximum height | $a = 9.81 \text{ m s}^{-2}$ downwards | Velocity is zero, acceleration is still $g$ |

## What's next

Free fall is a core application of constant acceleration kinematics, and forms the foundation for understanding projectile motion, where motion is split into constant velocity horizontal motion and constant acceleration vertical free fall motion. Free fall concepts also appear frequently in practical assessment questions, with the experiment to measure $g$ being one of the most common practical topics in CIE A-Level Physics. Mastering free fall sign conventions and problem solving patterns will make more complex motion topics much easier to tackle.

- [Projectile Motion](https://www.owlsprep.com/study/cie-9702-u2-projectile-motion/)
- [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/)

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