Falling, swinging and rolling: height becomes speed

7 minAS Level5.2.25.2.4

AI-drafted, independently checked · not yet teacher-reviewed

By the end of this lesson

By the end you can link the two energy stores with $mg\Delta h = \frac{1}{2}mv^2$ to find a speed from a height (or the other way round), and explain why mass cancels — so a dropped feather and a dropped hammer would tie in a vacuum.

Key to the symbols

The main symbols used in this lesson:

Symbol	Meaning
$E_P$	gravitational potential energy
$E_K$	kinetic energy
$\Delta$	a change in, so $\Delta v$ means "the change in $v$ "

A trade, not a loss

Lift a ball, then let go. Nothing is pushing it — gravity is doing the work now.

On the way down, the gravitational store empties and the kinetic store fills at the same rate. At every moment, ignoring air resistance,

$\text{loss in } E_P = \text{gain in } E_K$

At the bottom, the height's worth of energy has all become speed. That is the whole idea of this lesson: one equation, two stores.

Do the algebra once, then reuse it

Start with the exchange of energy for an object falling from rest through a height $h$ :

$mgh = \frac{1}{2}mv^2$

The mass $m$ appears on both sides, so divide both sides by $m$ :

$gh = \frac{1}{2}v^2$

Rearrange for $v$ :

$v = \sqrt{2gh}$

$m$ is gone. That is a strong statement: the final speed depends only on the height fallen and on $g$ — not on the mass, not on the shape.

The reason is that a heavier object has more gravitational energy available and needs more energy to reach a given speed, and the two effects cancel exactly.

Worked example: Worked example: dropping from a window

A ball is dropped from rest from a window $5.0\ \text{m}$ above the ground. Ignore air resistance.

Calculate its speed just before it lands.

Have a go first?

Quick check

CalculateQuestion 1 · Recall

Multiple choiceQuestion 2 · Recall

Answer the 3 questions above to continue.