Question

Cliff divers at Acapulco jump into the sea from a cliff 37.1 m high. At the level of the sea, a rock sticks out a horizontal distance of 13.39 m. The acceleration of gravity is 9.8 m/s 2 . With what minimum horizontal velocity must the cliff divers leave the top of the cliff if they are to miss the rock? Answer in units of m/s.

Answer 1

Answer:

$v_x = 4.87 m/s$

Explanation:

Height of the cliff is given as

$h = 37.1 m$

now the time taken by the diver to hit the surface is given as

$h = \frac{1}{2}gt^2$

$37.1 = \frac{1}{2}(9.8)t^2$

$t = 2.75 s$

Now in the same time it has to cover a distance of 13.39 m

so the speed in horizontal direction is given as

$v_x = \frac{x}{t}$

$v_x = \frac{13.39}{2.75}$

$v_x = 4.87 m/s$