For this let's take gravitational acceleration to be $10m/s^2$
So you're very high up. Like really high. You drop a stone vertically down. Then when that stone has descended 5m you release another one directly above the first.
Now the two stones are in freefall. Air resistance can be negligible for now.
Consider the distance between the two stones. Does it stay constant? If so, how much is that distance changing over time? And is that rate changing itself?
My maths teacher was wishing me a merry Christmas. I'm wondering if this problem needs calculus or not.
When the second stone is dropped, the first stone is traveling downwards with speed $10\text{ms}^{-1}$. Both stones then begin to accelerate downwards at the same rate so the speed that the second stone travels relative to the first remains constant. Thus the distance between the two stones is just $(5+10t)\text{m}$ where $t$ seconds have passed from the instant that the second stone is dropped. So the distance between the stones increases at a constant rate of $10\text{ms}^{-1}$.