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/s . Determine the time for the feather to fall to the surface of the moon .

Answer 1

Hi Hardeep..........

v^2=u^2+2.a.s   (where v=final velocity , u=initial velocity , s=distance covered , a=acceleration)
so here u=0 ; s=1.4 ; a=1.67
=> v^2 = 0+2*1.67*1.4
            =4.676
  => v = 2.16 m/sec
now,
v=u+at
=>  t = (v-u)/a
        =(2.16-0)/1.67
        =1.294 sec

Answer 2

Concept:
The gravitational pull on the surface of the moon is less than earth as the mass of the moon is lesser than the earth. We will find the time by using the expression  $s= ut + \frac{1}{2}at^{2}$ , where $s$ is the distance traveled, $t$ is the time taken, $u$ is the initial velocity, and $a$ is the acceleration.
Given:
Height from where the feather is dropped $= 1.4 m$
Acceleration due to gravity on the moon $= 1.67 ms^{-2}$
To Find:
Time taken by feather to fall on the surface of the moon.
Solution:
We know that, $s= ut + \frac{1}{2}at^{2}$
$\Rightarrow s = (0) \times t + \frac{1}{2} at^{2}$        [initial velocity $= 0$]
$\Rightarrow t^{2}= \frac {2\times s}{a}$
   $t^2= \frac{2\times1.4}{1.67}\\t^2= 1.6766\\t= \sqrt{1.6766}\\t= 1.29 seconds$
$\therefore$ tThe Time for the feather to fall on the surface of the moon is $1.29 seconds$