Question

A rock is projected upward from the surface of the moon, at time t = 0.0 s, with a velocity of 30 m/s. The acceleration due to gravity at the surface of the moon is 1.62 m/s2.
The height of the rock when it is descending with a velocity of 20 m/s is closest to

Answer 1

here,

initial velocity of rock, u = 30 m/s

acceleration due to gravity, g = 1.62 m/s

Time to reach maximum height:
t = u/g
t = 30/1.62
t = 18.519 s

From second eqn of motion :
height, h = u*t - 0.5*g*t^2
height, h = 30*18.519 - 0.5*1.62*18.519^2
height, h = 277.778 m

Time of fall :
t = 20 /1.62
t = 12.346 s

height durinh time of fall
h = 0 + 0.5 * g * t^2
h = 0.5 * 1.62 * 12.346^2
h = 123.463 m

Therefore ,

height during decend, H = 277.778 - 123.463

height during decend, H = 154.315 m