Answer
(a) W = -80.0 J
(b) W = -31.1 J
(c) Since the ball was moving at a slower average speed on the way down, the average force of air resistance was smaller on the way down. Therefore the magnitude of the work done by the air resistance was smaller.
Work Step by Step
(a) We can find the initial velocity $v_1$
$v_1 = \sqrt{(30.0~m/s)^2+(40.0~m/s)^2}$
$v_1 = 50.0~m/s$
$K_1+U_1+W = K_2+U_2$
$\frac{1}{2}mv_1^2+0+W = \frac{1}{2}mv_2^2+mgh$
$W = \frac{1}{2}mv_2^2+mgh-\frac{1}{2}mv_1^2$
$W = \frac{1}{2}(0.145~kg)(18.6~m/s)^2+(0.145~kg)(9.80~m/s^2)(53.6~m)-\frac{1}{2}(0.145~kg)(50.0~m/s)^2$
$W = -80.0~J$
(b) We can find the final velocity $v_3$
$v_3 = \sqrt{(11.9~m/s)^2+(-28.7~m/s)^2}$
$v_3 = 31.1~m/s$
$K_2+U_2+W = K_3+U_3$
$\frac{1}{2}mv_2^2+mgh+W = \frac{1}{2}mv_3^2+0$
$W = \frac{1}{2}mv_3^2-mgh-\frac{1}{2}mv_2^2$
$W = \frac{1}{2}(0.145~kg)(31.1~m/s)^2-(0.145~kg)(9.80~m/s^2)(53.6~m)-\frac{1}{2}(0.145~kg)(18.6~m/s)^2$
$W = -31.1~J$
(c) The magnitude of the work done by air resistance is smaller in part (b) than part (a) because the force of air resistance is greater when the speed of an object is greater. Since the ball was moving at a slower average speed on the way down, the average force of air resistance was smaller on the way down. Therefore the magnitude of the work done by the air resistance was smaller.
