$A$ raindrop of mass $1.00 \, g$ falling from a height of $1 \, km$ hits the ground with a speed of $50 \, m s^{-1}$. Calculate
$(a)$ the loss of $PE$ of the drop
$(b)$ the gain in $KE$ of the drop
$(c)$ Is the gain in $KE$ equal to loss of $PE$? If not,why?
Take $g = 10 \, m s^{-2}$.

(C) Given:
Mass $m = 1.00 \, g = 1.00 \times 10^{-3} \, kg$
Height $h = 1 \, km = 1000 \, m = 10^3 \, m$
Final velocity $v = 50 \, m s^{-1}$
Acceleration due to gravity $g = 10 \, m s^{-2}$
$(a)$ Loss in $PE = mgh = (1.00 \times 10^{-3} \, kg) \times (10 \, m s^{-2}) \times (10^3 \, m) = 10 \, J$.
$(b)$ Gain in $KE = \frac{1}{2}mv^2 = \frac{1}{2} \times (1.00 \times 10^{-3} \, kg) \times (50 \, m s^{-1})^2 = 0.5 \times 10^{-3} \times 2500 = 1.25 \, J$.
$(c)$ No,the gain in $KE$ is not equal to the loss in $PE$. This is because the raindrop experiences a resistive force (viscous drag) due to the air as it falls. $A$ significant portion of the potential energy is dissipated as heat due to the work done against this air resistance.