Constant acceleration equation along y:
v_{y}^{2} =
v_{oy}^{2} + 2a_{y} Δy

Multiply by m/2:
½ m v_{y}^{2} = ½ m v_{oy}^{2} + ma_{y} Δy

Example: a ball thrown straight up from y=0 and reaches y=h before falling back to y=0.

At top, final velocity is zero. We get:
0 = ½ m
v_{oy}^{2} - mgh or
mgh = ½ m v_{oy}^{2}

Going down, the initial velocity that is zero, so:
½ m v_{y}^{2} = mgh

*Obvious conclusion:* Final velocity
equals the initial velocity.

*Something new:* The kinetic energy
decreases on the way up, and increases back to the original amount on the way
down. The kinetic energy is transformed into
** gravitational potential energy**
U = mgh.

Kinetic energy is energy associated with motion.

Potential energy is energy associated with position.