Posted by Sean on .
a lunar landing module is descending to the moon's surface at a steady velocity of 10 m/s. At a height of 120m, a small object falls from its landing gear. Taking the moon's gravitational acceleration as 1.6m/s^2, at what speed, in m/s, does the object strike the moon?
Conservation of energy:
1/2 v^2 = 1/2 v_{0}^2 + *(1.6m/s^2)*120m
v_{0} = 10 m/s
The answer is suppose to Be 22
Let's see:
1/2 v^2 = 1/2 (10 m/s)^2 + (1.6m/s^2)*120m >
v^2 = (10 m/s)^2 + 2*(1.6m/s^2)*120m
v = sqrt[100 + 2*1.6*120] m/s=
22 m/s
OOO
i put 10m/s as the final velocity

Physics 
Zealot,
22