A stone is released from rest at a height of $20$ m. Neglecting air resistance, find its speed just before it hits the ground. (Take $g = 10$ m s$^{-2}$.)

By conservation of energy, $mgh = \dfrac{1}{2}mv^2$, so $v = \sqrt{2gh} = \sqrt{2 \times 10 \times 20} = \sqrt{400} = 20$ m s$^{-1}$. Omitting the factor 2 gives $\sqrt{200} \approx 14.1$.