A simple pendulum of length $l = 1$ m is displaced from the vertical by an angle $\alpha = 60°$ and released. Find the speed of the bob at the lowest point. Neglect air resistance; $g = 10$ m/s².
When displaced, the bob rises to a height
$$h = l(1 - \cos\alpha) = 1 \cdot (1 - \cos 60°) = 1 \cdot 0.5 = 0.5 \text{ m}$$
By conservation of energy $mgh = mv^2/2$, hence
$$v = \sqrt{2gh} = \sqrt{2 \cdot 10 \cdot 0.5} = \sqrt{10} \approx 3.16 \text{ m/s}$$
$v \approx 3.16$ m/s