Answer
The ball's greatest height above the ground is 8.61 meters.
Work Step by Step
$K_2+U_2 = K_1+U_1$
$\frac{1}{2}mv_2^2+mgh=\frac{1}{2}mv_1^2+0$
$v_2^2+2gh=v_1^2$
$h=\frac{v_1^2-v_2^2}{2g}$
$h=\frac{(15~m/s)^2-(15~m/s~cos(60.0^{\circ}))^2}{(2)(9.80~m/s^2)}$
$h = 8.61~m$
The ball's greatest height above the ground is 8.61 meters.
