A feather is dropped on the moon from a height of 1.40 meters. The acceleration of gravity on the moon is 1.67 m/s2. Determine the time for the feather to fall to the surface of the moon. Solution Given: vi = 0 m/s d = -1.40 m a = -1.67 m/s2 Find: t = ?? d = vi*t + 0.5*a*t2 - 1.40 m = (0 m/s)*(t)+ 0.5*(-1.67 m/s2)*(t)2 -1.40 m = 0+ (-0.835 m/s2)*(t)2 (-1.40 m)/(-0.835 m/s2) = t2 1.68 s2 = t2 t = 1.29 s.

1. A feather is dropped on the moon from a height of 1.40 meters. The acceleration of gravity on
the moon is 1.67 m/s2. Determine the time for the feather to fall to the surface of the moon.
Solution
Given: vi = 0 m/s d = -1.40 m a = -1.67 m/s2 Find: t = ?? d = vi*t + 0.5*a*t2 -
1.40 m = (0 m/s)*(t)+ 0.5*(-1.67 m/s2)*(t)2 -1.40 m = 0+ (-0.835 m/s2)*(t)2 (-1.40 m)/(-0.835
m/s2) = t2 1.68 s2 = t2 t = 1.29 s