a person standing at the edge of a seaside cliff kicks a stone over the edge with a speed of 18 ms. The cliff is 52m above the waters surface. How long does it take for the stone to fall to the water? With what speed does it strike the water?

Answer 1

Since the stone start with no initial vertical velocity component, the time T to hit the water is given by

H = 52 m = (1/2) g T^2
which leads to
T = sqrt(2H/g)

The speed V2 when it hits the surface is most easily computed using energy conservation.

(1/2) m V2^2 = (1/2) m V1^2 + m g H
V2^2 = V1^2 + 2gH

V1 is the initial velocity, 18 m/s.
You know what g is, I assume.

Answer 2

e=mc^2