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

(50 M) Let the initial heights be $H_1 = 150 \, m$ and $H_2 = 100 \, m$.
Both objects are dropped from rest,so their initial velocity $u = 0 \, m/s$.
Both objects fall with the same acceleration due to gravity,$g$.
Distance covered by an object in time $t$ is given by $s = ut + \frac{1}{2}gt^2$.
Since $u = 0$,the distance covered by both objects in $t = 2 \, s$ is $s = \frac{1}{2}g(2)^2 = 2g$.
After $2 \, s$,the height of the first object from the ground is $h_1 = H_1 - s = 150 - 2g$.
The height of the second object from the ground is $h_2 = H_2 - s = 100 - 2g$.
The difference in their heights after $2 \, s$ is $h_1 - h_2 = (150 - 2g) - (100 - 2g) = 150 - 100 = 50 \, m$.
Since the acceleration $g$ is the same for both,the distance covered by both objects at any time $t$ is the same $(s = \frac{1}{2}gt^2)$.
Therefore,the difference in heights at any time $t$ is $(H_1 - s) - (H_2 - s) = H_1 - H_2 = 150 - 100 = 50 \, m$.
Thus,the difference in heights remains constant and does not vary with time.