A body is dropped from rest at a height of `150 m` and simultanceously, another body is dropped from rest from a point `100 m` above the ground. What is the difference between heights after theymave fallen for (i) `3s (ii) 5s`. Consider that the body on reaching fround remains there and acceleration due to gravity be `10 m//s^(2)`.

Explanation

(i) For the first body taking downward motion of the body, we get have
`u=0, t=3 s, a=10 m//s^(2), S=?`
As, S=ut +1/2 at^(2)`
:. `S=0 xx 3+ 1/2 xx 10 xx 3^(2) =45 m`
Height of body above ground `=150 -45`
`=105 m`.
For second body
Downward distance travelled in `3 seconds`
`=45 m`
Height of bodyablve ground `=100 -45`
`=55 m`
Thus,differnce in heights `=105-55=50 m`
(ii0 Similarly distance coverd by each body in `5 second will be given by
`S=1/2 gt^(20 =1/2 xx 10 5^(2) =125 m`
Now, the height of first body above ground
`=150-125 =25 m`
For second body , as the distance travelled is more than its height from ground, it means this body will be on ground just after `5 seconds. Therfore, the height of this body from ground is zero.
Thus, differnce in height of two bodies
`=25 -0 25 m`.