# a) Show that P(h0)=K(vf) b) Interpret result from part a) Given: N

a) Show that $P\left(h0\right)=K\left(vf\right)$
b) Interpret result from part a)
Given:
A ball with mass m is dropped from an initial height of ${h}_{0}$, and lands with a final velocity of vf. The kinetic energy of the ball is $K\left(v\right)=12mv2$, where v is its velocity and the potential energy of the ball is $P\left(h\right)=mgh$, where h is its height and g is a constant.
$vf=2gh0$
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Calculation:
a) The kinetic energy of the ball at final velocity is
$K\left(vf\right)=K\left(2gh0\right)$
$K\left(vf\right)=12m\left(2gh0\right)2$
$K\left(vf\right)=12m\left(2gh0\right)$
$K\left(vf\right)=mgh0$
The potential energy at ${h}_{0}$ is $P\left(h0\right)=mgh0$
Therefore $P\left(h0\right)=K\left(vf\right)$
b) The potential energy of the ball before it is dropped is equal to the kinetic energy of the ball when it lands.
Conclusion:
c) $P\left(h0\right)=K\left(vf\right)$
d) The potential energy of the ball before it is dropped is equal to the kinetic energy of the ball when it lands.