A ball is dropped from a height of 24 m.

Step 1. $mgh = \tfrac{1}{2}mv^2$, so $v = \sqrt{2gh}$.

Step 2. $v = \sqrt{2 \times 9.8\ \text{m/s}^2 \times 24\ \text{m}} = \sqrt{470.4\ \text{m}^2/\text{s}^2} = 21.69\ \text{m/s}$.