Consider a drop of rain water having mass 1g falling from a height of 1km. if it hits the ground with a speed of $50{\text{m/s}}^{\text{2}}$ then work done by the gravitational force and resistance force of air is $\text{(take}\text{g}=1{\text{0m/s}}^{\text{2}})$

$\text{initial}\text{energy}=\text{mgh}+\text{K}\text{.E}$

$={10}^{-3}\times 10\times {10}^3=10\text{J}$

But it is given to be reaching the earth with$\text{v}=5\text{0m/s}$

$\text{K}\text{.E}=\frac{1}{2}\times {\text{mv}}^{\text{2}}=\frac{1}{2}{10}^{-3}\times 50\times 50$

$=1.25\text{J}$

Hence the work done by resisting force due air$=\text{final}\text{energy}-\text{initial}\text{energy}$

$=1.25-10$

$\Rightarrow -8.75\text{J}$