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 = 21 s to return to the surface.
Write an expression for the maximum height achieved by the ball on the moon, hm, in terms of the gravitational acceleration on earth, g, and the time, t.

Answer 1

$t=t_1+t_2,t_1=t_2\to t=2t_2\to t_2=t/2$

$h_m=\frac{g_Mt_2^2}{2}=\frac{(g/6)(t/2)^2}{2}=\frac{gt^2}{48}=\frac{9.81\cdot21^2}{48}\approx90.1(m)$