Past exam of the medicine course of the University of Oxford / 1867 / First examination / Physics / 3 / Solution

Take downward displacement as positive and gravitational acceleration $g$ as constant. In a vacuum, $mdv/dt=mg$, so with initial downward velocity $u$,For release from rest these become $v=gt$ and $s=g{t}^2/2$.

A raindrop encounters an increasing upward drag force as its speed grows. At terminal velocity, gravitational weight is balanced by drag force and the small buoyancy contribution:The drop remains in motion: zero acceleration does not imply zero velocity. In a quadratic drag force model, $F_{\mathrm{drag}}={\rho }_{\mathrm{air}}{C}_{D}A{v}^2/2$, giving $v_t=\sqrt{2(m-{\rho }_{\mathrm{air}}V)g/({\rho }_{\mathrm{air}}{C}_{D}A)}$. That formula assumes a fixed effective area and drag coefficient; sufficiently small drops or deforming large drops need an appropriate model. The question's “effective force” at settled terminal velocity is the approximately zero resultant force, not its nonzero gravitational weight.