A rock is thrown upward with a velocity of 20 meters per second from the top of a 37 meter high cliff, and it misses the cliff on the way back down. When will the rock be 4 meters from the water, below? Round your answer to two decimal places.

Answer 1

Let us name
vr=velocity of the rock in the moment of throwing ( in m/s )

v0=velocity 0 m/s

vw=velocity of the rock 4 meters from the water.

hmax= top height of the rock measured from the water level

h= the height of the rock measured from the surface of the cliff.

hc= the height of the cliff

ha= the distance from the highest point the rock reaches by being thrown to the 4 meters depth of water.

Aplying Galilei's formula we have:
vr^2=2*g*h
so h=(vr^2)/2g=400/20=20m

But hmax=hc+h=37+20=57m
ha=57-(-4)=61m

So vw=sqrt(2*g*h)=34,93 m/s