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?

Question

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

Answer 1

, but it should be 22m/s