Question

A projectile is launched vertically from the surface of the Moon with an initial speed of 1100 m/s. At what altitude is the projectile's speed one-fourth its initial value?

Answer 1

Given

   projectile launched vertiallcy from surface of Moon where

acceleration due to gravity g = 1.633 m/s2

   with initial velocity u = 1100 m/s

and the final velocity v = u/4

   height h = ?

from equations of motion

       v^2 - u^2 = - 2*g*h

       h = v^2 - u^2 / (-2*g)

substituting
       h = (1/4 u)^2 -u^2 /(-2*g)
       h = u^2(1/16 -1) /2g
       h = 1100^2(-15/16) /(-2*1.633) m
  
       h = 347328.53644 m

so the altitude where the velocity becomes 1/4 th of initial velocity is 347328.53644 m