Answer
(a) The rock's speed just after it left the ground is 30.3 m/s.
(b) The maximum height is 46.8 meters.
Work Step by Step
(a) $KE_1 + W = KE_2$
$\frac{1}{2}mv_1^2 - mgh = \frac{1}{2}mv_2^2$
$v_1^2 = v_2^2+2gh$
$v_1 = \sqrt{v_2^2+2gh}$
$v_1 = \sqrt{(25.0~m/s)^2+(2)(9.80~m/s^2)(15.0~m)}$
$v_1 = 30.3~m/s$
The rock's speed just after it left the ground is 30.3 m/s.
(b) $KE_1 + W = KE_2$
$\frac{1}{2}mv_1^2 - mgh = 0$
$\frac{1}{2}v_1^2 = gh$
$h = \frac{v_1^2}{2g}$
$h = \frac{(30.3~m/s)^2}{(2)(9.80~m/s^2)}$
$h = 46.8~m$
The maximum height is 46.8 meters.
