An object is dropped from rest at a height of `150 m` and simultanously another object is dropped from rest at a height `100m`. What is the difference in their height after `2 s` if both the objects drop with same acceleration ? How does the difference in height vary with time ?

Explanation

For first object given, u = 0 (because object dropped from rest) and time (t) = 2s.
From second equation of motion, the distance covered by first object in 2s is

`h = ut +1 /2g t^(2)`
h = `0 xx2 + 1/2xx10xx(2)^(2) [because g = 10m//s^(2)]`
h = `0+1/2xx10xx4 = 20m`
Height of first object from the ground after 2 s `(h_(1))` = 150m - 20m = 130m
For second object , u = 0 and time (t) = 2s
From second equation of motion , the distance by second object in 2 s is
h = ut `+1/2g t^(2) = 0xx2 + 1/2xx10xx(2)^(2) [because g = 10m//s^(2)]`
= 0 `+1/2xx10xx4 = 20m`
Height of second object from the ground after 2s then `h_(2)= 100m - 20m = 80m`
Now , difference in the height after 2s = `h_(1)-h_(2) = 130 - 80 = 50m`
The difference in hights of the objects will remain same with time as both the object have been dropped from rest and are falling with same acceleration i.e., (acceleration due to gravity).