A rain drop of radius 2mm falls from a height of 500 m above the ground. It falls with decreasing acceleration (due to viscous resistance of the air) until at half its original height, it attains its maximum (terminal) speed, and moves with uniform speed thereafter. What is the work done by the gravitational force on the drop in the first and second half of its speed on reaching the ground is `10 ms^(-1)`?

Explanation

`0.082 J` in each half , `-0.163 J`