## Problem Statement

**A feather is dropped on the moon from a height of 10 meters. The acceleration of gravity on the moon is one sixth of that on the earth. Determine the time for the feather to fall to the surface of the moon. **

## Solution

**Acceleration of gravity on the moon = 10/6 m/s ^{2} = 5/3 m/s^{2} = g_{moon}**

To find out the time we will use this eqn. V = U + g_{moon}.t

As it falls from rest, so U = 0.

So the equation changes to, V = g_{moon}.t

So, time t = V/ g_{moon } … (1)

Now we have to find out the value of V.

**V ^{2} = U^{2} + 2 g_{moon} h**

Considering U = 0, we get V =

**√**

**(2 g**

_{moon}h) =**√**

**((5/3)*10) = 4 m/s (approx) …(2)**

so from eqn 1, the time t = V/ g

so from eqn 1, the time t = V/ g

_{moon }= 4/(5/3)= 12/5 = 2.4 secSolution to problems – class 9 – Set 1 Q 21