Answer
(a) v = 24.0 m/s
(b) v = 24.0 m/s
(c) With air resistance, part (b) would give a higher speed than part (a).
Work Step by Step
(a) $K_2 = K_1 + U_{grav}$
$\frac{1}{2}mv_2^2 = \frac{1}{2}mv_1^2+mgh$
$v_2^2 = v_1^2+2gh$
$v_2 = \sqrt{v_1^2+2gh}$
$v_2 = \sqrt{(12.0~m/s)^2+(2)(9.80~m/s^2)(22.0~m)}$
$v_2 = 24.0~m/s$
(b) We can do the same calculation as part (a) to find that the speed of the ball just before striking the ground is 24.0 m/s
(c) Since the path of the ball in part (a) follows a longer distance, the opposing force of air resistance will remove more energy from the ball. Thus, with air resistance, part (b) would give a higher speed than part (a).
