Answer to Question #195884 in Physics for Mary

Question

Nathan was challenged to dive from a cliff. Starting from rest, he jumped off a cliff that is 9.0 m above the water. Solve the following: Show your solution

A. How long would it takd Nathan to reach the water?

B. How fast would his velocity be before reaching the water if the effect of air resistance is neglected?

Answer 1

A)

$$y=\dfrac{1}{2}gt^2,$$ $$t=\sqrt{\dfrac{2y}{g}},$$ $$t=\sqrt{\dfrac{2\cdot9.0\ m}{9.8\ \dfrac{m}{s^2}}}=1.35\ s.$$

B)

$$v=-gt=-9.8\ \dfrac{m}{s^2}\cdot1.35\ s=-13.2\ \dfrac{m}{s}.$$

The sign minus means that Nathan's velocity directed downward.