Question

Suppose an astronaut drops a feather from 1.2 m above the surface of the Moon. If the acceleration due to gravity on the Moon is 1.62 m/s2 downward, how long does it take the feather to hit the Moon's surface? s

Answer 1

This is a freefall movement, which is unformly accelerated.

Then the equation that rules the relation of distance and time is:

d = Vo*t + at^2/2

Where Vo, is the initial velocity =0
a = acceleration = 1.62 m/s^2
t = time
d= distance

Then, d = a*t^2/2 ==> t = sqrt(2d/a) = sqrt (2*1.2m/1.62m/s^2) = 1.2 sec.