Answer to Question #159572 in Physics for Pat

Question

On the moon, the acceleration due to gravity is 1/6 of Earth's. A ball is thrown straight up on the moon and it takes t = 29 s to return to the surface. What is the difference between the maximum height of the ball thrown on the moon and the ball being thrown on earth?

Answer 1

In all equations considered here, g is that of the Earth.

The maximum height of a body thrown vertically upward is related to the acceleration due to gravity and total time of flight like follows:

$$h=\frac{gt^2}{8}.$$

This height corresponds to the initial vertical velocity of

$$v=\sqrt{2gh}=\frac{gt}{2}.$$

Thus, for the Moon this is

$$v=\frac{gt_M}{12},\\\space\\ h_M=\frac{gt_M^2}{48}$$

Therefore, the maximum height of the ball thrown at the same velocity on the Earth will be

$$h_E=\frac{v^2}{2g}=\frac{gt_M^2}{288}.$$

The difference between the maximum height of the ball thrown on the moon and the ball being thrown on earth is

$$\Delta h=h_M-h_E=\frac{gt_M^2}{48}-\frac{gt_M^2}{288}=\\\space\\ =\frac{5gt_M^2}{288}=\frac{5\cdot9.8\cdot29^2}{288}=143\text{ m}.$$