A ball of mass 0.5 kg is released from rest at a height of 30 m. How fast is it going when it hits the ground? Acceleration due to gravity is g = 9.8 m/s2.

Assume you don't consider air friction. Then mgh = 1/2 mv^2 v = sqrt(2gh) = sqrt(2*9.8 * 30) = 24.25 m/s

[latex]Given:\\m=0.5kg\\h=30m\\g=9.8 \frac{m}{s^2} \\\\Find:\\v=?\\\\Solution:\\\\E_p=E_k\\\\mgh= \frac{mv^2}{2} \\\\2gh=v^2\Rightarrow v= \sqrt{2gh}\\\\v= \sqrt{2\cdot 9.8 \frac{m}{s^2} \cdot 30m} \approx24.25 \frac{m}{s} [/latex]