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...

A rock launched with an initial speed will move outward from the moon's surface (that is, it will go up). Its motion will be resisted by the gravity of the moon, which will cause deceleration and ultimately the rock will stop and then fall back. The point where the velocity of the rock (during its upward journey) is zero is also its maximum height.

Thus, at the maximum height velocity = 0 m/s.

Initial velocity, u = 30 m/s

final velocity, v = 0 m/s

acceleration, a = 1.62 m/s^2 (in downward direction)

Using the equation of motion, v^2 = u^2 + 2as

we get, 0^2 = 30^2 + 2 x -1.62 x s

or, s = 30^2 / (2 x 1.62) 

solving this equation, we get: s = 277.78 m. 

Thus, the rock will achieve a maximum height of 277.78 m before falling back to the moon's surface.

Hope this helps.