Ask Question
5 December, 14:41

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

+5
Answers (1)
  1. 5 December, 14:51
    0
    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.
Know the Answer?
Not Sure About the Answer?
Find an answer to your 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 ...” in 📗 Physics if the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers