#### Answer

(a) The maximum height is 25 meters above the ground.
(b) The speed of the ball is 10 m/s
(c) The speed at impact is 22 m/s

#### Work Step by Step

(a) Let $KE_1$ be the initial kinetic energy. Let $PE_1$ be the initial potential energy. Let $KE_2$ be the kinetic energy at maximum height. Let $PE_2$ be the potential energy at maximum height.
We can use conservation of energy to find the maximum height $h_2$.
$KE_2+PE_2 = KE_1+PE_1$
$0+mg~h_2 = \frac{1}{2}mv_1^2+mg~h_1$
$h_2 = \frac{\frac{1}{2}v_1^2+g~h_1}{g}$
$h_2 = \frac{\frac{1}{2}(10~m/s)^2+(9.80~m/s^2)(20~m)}{9.80~m/s^2}$
$h_2 = 25~m$
The maximum height is 25 meters above the ground.
(b) Let $KE_3$ be the kinetic energy as the ball is passing the window on the way down. Let $PE_3$ be the potential energy at the height of the window.
We can use conservation of energy to find the speed $v_3$.
$KE_3+PE_3 = KE_1+PE_1$
Since $PE_3=PE_1$, then $KE_3 = KE_1$. Therefore, $v_3 = v_1$. The speed of the ball is 10 m/s
(c) Let $KE_4$ be the kinetic energy at the moment of impact. Let $PE_4$ be the potential energy at ground level.
We can use conservation of energy to find the speed at impact $v_4$.
$KE_4+PE_4 = KE_1+PE_1$
$\frac{1}{2}mv_4^2+0 = \frac{1}{2}mv_1^2+mg~h_1$
$v_4^2 = v_1^2+2g~h_1$
$v_4 = \sqrt{v_1^2+2g~h_1}$
$v_4 = \sqrt{(10~m/s)^2+(2)(9.80~m/s^2)(20~m)}$
$v_4 = 22~m/s$
The speed at impact is 22 m/s