Question

A raindrop of mass $m$ at a height $h$ starts from rest and initially accelerates under gravity. Due to viscous drag of air, its motion is resisted and it finally it acquires a constant terminal velocity $v_T$ and hits the ground. Neglect the buoyancy, assume acceleration due to gravity $g$ and no wind. Find the magnitude of the average viscous force on the rain drop during the entire downward trip.

Answer 1

$P_A=P_B{P}_0+d_1(\cos \alpha -\sin \alpha )=P_0+d_2(\cos \alpha +\sin \alpha )\frac{d_1}{d_2}=\frac{\cos \alpha +\sin \alpha }{\cos \alpha -\sin \alpha }$

dividing by $\cos \alpha$ to both numerator and denominator we will get,

$\frac{d_1}{d_2}=\frac{1+\tan \alpha }{1-\tan \alpha }$