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29 July, 13:10

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

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  1. 29 July, 13:24
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    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.
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