Question

A ski lift is used to transport people from the base of a hill to the top. If the lift leaves the

Answer 1

The energy of the ski lift at the base is kinetic energy:
$K= \frac{1}{2}mv^2$
where m is the mass of the ski lift+the people carried, and $v=15.5 m/s$ is velocity at the base.
As long as the ski lift goes upward, its velocity decreases and its kinetic energy converts into potential energy. Eventually, when it reaches the top, its final velocity is v=0, so no kinetic energy is left and it has all converted into gravitational potential energy, which is 
$U=mgh$
where $g=9.81 m/s^2$ and h is the height at the top of the hill.

So, since the total energy must conserve, we have
$U=K$
and so
$mgh = \frac{1}{2}mv^2$
from which we find the height:
$h= \frac{v^2}{2g}= \frac{(15.5m/s)^2}{2\cdot 9.81 m/s^2}=12 m$