Answer to Question #157963 in Mechanics | Relativity for John Bittinger

Question

A Cliff Calamity is going to attempt a dive into the surf. This no simple dive, but

a cliff dive from the maximum allowable competitive height of 28 m (about 92

feet). Complicating the dive is some rocks in the surf for about 7 m horizontally

from the cliff face. Cliff takes off from the cliff with a vertical velocity of 0 m/s

and a horizontal velocity of 3.5 m/s. Does cliff clear the rocks? What is Cliff’s

vertical velocity at the instant of impact?

Answer 1

a) Let's first find the time that Cliff Calamity takes to reach the water:

$$y=\dfrac{1}{2}gt^2,$$ $$t=\sqrt{\dfrac{2y}{g}}=\sqrt{\dfrac{2\cdot28\ m}{9.8\ \dfrac{m}{s^2}}}=2.4\ s.$$

Then, we can find the horizontal displacement of Cliff Calamity:

$$x=v_0t=3.5\ \dfrac{m}{s}\cdot2.4\ s=8.4\ m.$$

As we can see from the calculations, Cliff clear the rocks.

b) The vertical component of the Cliff's velocity at the instant of impact can be found as follows:

$$v_y=v_{0y}-gt=0-9.8\ \dfrac{m}{s}\cdot2.4\ s=-23.52\ \dfrac{m}{s}.$$

The sign minus means that the vertical velocity directed downward.