A 2-kg rock is thrown vertically upward at a speed of 3.2 m/s from the surface of the moon. If it returns to its starting point in 4 seconds, what is the acceleration due to gravity on the moon?

Let [tex]g_m[/tex] be the gravitational acceleration of the moon. We know that due to the law of energy conservation, kinetic energy (and speed) of the rock when being thrown upwards from the surface and when it returns to the surface is the same. Given that [tex]g_m[/tex] stays constant, we can conclude that the time it takes to reach its highest point, aka 0 velocity, is the same as the time it takes to fall down from that point to the surface, which is half of the total time, or 4 / 2 = 2 seconds.

So essentially it takes 2s to decelerate from 3.2 m/s to 0. We can use this information to calculate [tex]g_m[/tex]

[tex]g_m = \frac{\Delta v}{\Delta t} = \frac{0 - 3.2}{2} = \frac{-3.2}{2} = -1.6 m/s^2[/tex]

So the gravitational acceleration on the Moon is 1.6 m/s2