Question

A feather is dropped onto the surface of the moon. How far will the feather have fallen if it reaches the surface in 9.00 seconds?
Given: g on moon = -1.63 meters/second2

Answer 1

Take in count that in the moon there is no air resistance. The feather will fall as it were a hammer (or any other heavy object). Now:
\[h = V_0t + \frac{ 1 }{ 2 } g_{moon}t^2\]
The feather is dropped, so Vo = 0.
\[h = \frac{ 1 }{ 2 } g_{moon}t^2\]
\[h = \frac{ 1 }{ 2 } (1.63 m/s^2) (9s)^2\]
\[h = 66 m\]

Answer 2

:)