Know that the acceleration of free fall for an object near to the Earth is approximately constant and that this is equivalent to the acceleration of free fall

1.2 Motion – Acceleration of Free Fall

Learning Objective

Know that the acceleration of free fall for an object near the Earth is approximately constant and that this constant value is denoted by g.

Key Concepts

• When an object moves only under the influence of Earth's gravity, its acceleration is called the acceleration of free fall.
• Near the Earth's surface, this acceleration is essentially constant and is denoted by g.
• The accepted numerical value for g in IGCSE examinations is 9.8 m s⁻² (often rounded to 10 m s⁻² for quick calculations).
• Direction of g is vertically downwards.

Why is g Approximately Constant?

Earth’s radius is about 6.4 × 10⁶ m. For heights h that are small compared with this radius (e.g., less than a few kilometres), the change in distance from the centre of the Earth is negligible, so the gravitational force – and therefore the acceleration – changes very little. Hence we treat g as constant.

Mathematical Description

The acceleration of a freely falling object is expressed as:

\$a = g = 9.8\ \text{m s}^{-2}\$

When using the rounded value:

\$a \approx 10\ \text{m s}^{-2}\$

Equations of Motion with Constant Acceleration

For motion under constant acceleration (including free fall), the following kinematic equations apply:

1. \$v = u + at\$
2. \$s = ut + \frac{1}{2}at^{2}\$
3. \$v^{2} = u^{2} + 2as\$

Where:

• \$u\$ = initial velocity (m s⁻¹)
• \$v\$ = final velocity (m s⁻¹)
• \$a\$ = acceleration (m s⁻²) – for free fall \$a = g\$
• \$t\$ = time (s)
• \$s\$ = displacement (m)

Typical Situations

Sample Calculation

Calculate the time it takes for a stone dropped from a height of 20 m to reach the ground. Use \$g = 10\ \text{m s}^{-2}\$ for simplicity.

Given:

• \$u = 0\$ (dropped from rest)
• \$s = 20\ \text{m}\$ (downwards)
• \$a = g = 10\ \text{m s}^{-2}\$

Use \$s = ut + \frac{1}{2}gt^{2}\$:

\$20 = 0 \times t + \frac{1}{2}(10)t^{2}\$

\$20 = 5t^{2}\$

\$t^{2} = 4\$

\$t = 2\ \text{s}\$

Therefore, the stone reaches the ground after 2 seconds.

Common Misconceptions

• “Free fall means falling straight down.” – Any object moving only under gravity, even if initially projected upwards or downwards, is in free fall.
• “The acceleration changes as the object speeds up.” – Acceleration due to gravity remains constant; only the velocity changes.
• “Air resistance can be ignored always.” – In IGCSE questions unless stated otherwise, air resistance is ignored, but in real life it reduces the net acceleration.

Practice Questions

1. A ball is thrown vertically upwards with an initial speed of 15 m s⁻¹. Using \$g = 10\ \text{m s}^{-2}\$, calculate:

   1. The time taken to reach its highest point.
   2. The maximum height above the launch point.

2. An object is dropped from a height of 45 m. Determine the speed just before it hits the ground (use \$g = 9.8\ \text{m s}^{-2}\$).
3. A stone is thrown downwards from a cliff with an initial speed of 5 m s⁻¹. If it hits the water 3 s later, find the height of the cliff (use \$g = 10\ \text{m s}^{-2}\$).

Suggested Diagram

Summary

The acceleration of free fall, g, is a constant value (≈ 9.8 m s⁻²) for objects moving only under Earth’s gravity near the surface. This constancy allows the use of simple kinematic equations to solve a wide range of motion problems in the IGCSE Physics syllabus.