Cliff divers at Acapulco jump into the sea

from a cliff 31.5 m high. At the level of the

sea, a rock sticks out a horizontal distance of

12.49 m.

The acceleration of gravity is 9.8 m/s.

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.

Question

Jon Cline

Physics

11 July, 03:04

Cliff divers at Acapulco jump into the sea

from a cliff 31.5 m high. At the level of the

sea, a rock sticks out a horizontal distance of

12.49 m.

The acceleration of gravity is 9.8 m/s.

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.

+4

Answers (1)

Know the Answer?

Salvatore May · Answer 1

V=4.927 m/s

Explanation:

H=Vi*t+0.5*g*t² ... 1

initial velocity is Vi=0 m/s

H=31.5

putting value of H in equation 1

31.5=0+0.5*9.81*t²

t=2.535 s

S=V*t ... 2

Horizontal distance S=12.49 m

12.49=V * (2.535)

V=4.927 m/s