Question

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.

Answer 1

The law of motion of the object falling from height $H$is $y(t) = H - \frac{g t^2}{2}$, where $g$ is the acceleration of the gravity. When the object hits the surface $y(t) = 0$, from where $t = \sqrt{\frac{2 H}{g}}$.

Substituting $H = 1.4 m$ and $g = 1.67 \frac{m}{s^2}$ into the last formula, obtain $t \approx 1.3 s$.