Suppose an austronaunt drops a feather from 1.2 m above the surface of the moon if the acceleration due to?

....gravity on the moon is 1.62m/s^2 downward, how long does it take the feather to hit the Moon's surface?Suppose an austronaunt drops a feather from 1.2 m above the surface of the moon if the acceleration due to?
Ignoring air resistance (since it's the moon)





distance fallen from rest is given by:


d = 1/2 at^2





so t = sqrt (2d / a). They give you d and a. Plugnchug.Suppose an austronaunt drops a feather from 1.2 m above the surface of the moon if the acceleration due to?
1.217 seconds (to three decimal places)





Time=square root of Twice Distance travelled over acceleration





Where starting velocity is zero, of course.
1.21716 seconds.
from


s = s0 + v0*t - 1/2gt^2


we know


s0 = 0


s = -1.2


v0=0





so solve for t





t = sqrt(2*1.2/1.62)


t = 1.22 s