Physics
posted by Trisha .
A person standing at the edge of a seaside cliff kicks a stone over the edge with a speed of vi = 20 m/s. The cliff is h = 53 m above the water's surface, as shown below.
(a) How long does it take for the stone to fall to the water?
(b) With what speed does it strike the water?

Since there is no initial vertical velocity component, the time to fall (T) is given by
H = (1/2)gT^2
H is the cliff height.
T = sqrt(2H/g)
The speed Vf at impact can be easily computed with conservation of energy.
(1/2)m Vf^2 =(1/2)m Vo^2 = mgH
The mass cancels out. Vo is the initial velocity (when kicked)
Vf^2 = Vo^2 + 2 g H 
Oh i get it. You have to use the equation for an accelerating straightline object. Thanks