Question

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.

Answer 1

V=4.927 m/s

Step-by-step 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