Question

A 12.0 kg rock slides off the edge of a bridge and falls into the water 25.0 meters below. what is the kinetic energy of the rock just as it hits the water?

Answer 1

During the fall, all the initial potential energy of the rock 
$U=mgh$
has converted into kinetic energy of motion
$K= \frac{1}{2}mv^2$
where h is the initial height of the rock, m its mass, and v its velocity just before hitting the water. So, for energy conservation, we have
$U=K$
and so we can find the value of K, the kinetic energy of the rock just before hitting the ground:
$K=U=mgh = (12.0 kg)(9.81 m/s^2)(25.0 m)=2943 J$