Answer to Question #97712 in Physics for Michael

Question

A man steps off an overhanging cliff and falls 9m into a lake. How long does it take him to hit the water? How fast is he falling when he reaches the lake’s surface?

Answer 1

The equations of motion of a man are $y(t) = h - \frac{g t^2}{2}$ and $v(t) = gt$, where $h$ is the height of the cliff, $g = 9.81 \frac{m}{s^2}$- gravitational acceleration. Equating $y(t)$ to zero, and solving for $t$, obtain $0 = h - \frac{g t^2}{2} \Rightarrow t = \sqrt{\frac{2 h}{g}}$ - this is the time it takes to fall down. Calculating speed at that moment, obtain $v = g \sqrt{\frac{2 h}{g}} = \sqrt{2 g h}$.

Calculating for $h = 9 m$, obtain $t \approx 1.35 s$, $v \approx 13.29 \frac{m}{s}$.