Answer
The kinetic energy of the clown as he lands in the net is $~~5400~J$
Work Step by Step
We can use conservation of energy to find $K_2$, the kinetic energy of the clown as he lands in the net:
$K_2+U_2 = K_1+U_1$
$K_2 = K_1+U_1-U_2$
$K_2 = \frac{1}{2}mv_1^2+mgh_1-mgh_2$
$K_2 = \frac{1}{2}mv_1^2+mg(h_1-h_2)$
$K_2 = \frac{1}{2}(60~kg)(16~m/s)^2+(60~kg)(9.8~m/s^2)(-3.9~m)$
$K_2 = 5400~J$
The kinetic energy of the clown as he lands in the net is $~~5400~J$
