Question

A body falling freely, under the action of gravity passes two points 10 metres apart
vertically in 0.2 second. From what height, above the higher point, did it start to fall ?

Answer 1

Answer:

10.2hight it will be fall

Answer 2

Given:

A body falling freely, under the action of gravity passes two points 10 meters apart vertically in 0.2 seconds.

To Find:

Height above the higher point from where it started falling.

Solution:

We've assumed downward direction to be positive for this problem.

We know that a freely falling object falls under uniform acceleration, $g=9.8m/s^{2}$ downwards.

Let us assume that the body starts falling from a height $h$ above the higher point.

So, the initial velocity of the body, $u=0m/s$

Let the velocity at the higher point be $v$.

Using equations of motion

     $s=vt+\frac{1}{2} gt^2$

⇒  $10=v(0.2)+\frac{1}{2}(9.8)(0.2)^2$

⇒  $10=0.2v+0.196$

⇒  $v=49.02m/s$

Now,

     $v^2-u^2=2gh$

⇒  $(49.02)^2=2*9.8*h$

⇒  $h=122.6m$

Hence, the body started falling 122.6 meters above the higher point.