= Solution
Take downward <displacement> as positive and <gravitational acceleration> $g$ as constant. In a vacuum, $m\,dv/dt=mg$, so with initial downward <velocity> $u$,
$$
\boxed{v=u+gt,\qquad s=ut+\tfrac12gt^2.}
$$
For release from rest these become $v=gt$ and $s=gt^2/2$.
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:
$$
\boxed{F_{\rm net}=mg-F_{\rm drag}-F_{\rm buoyancy}=0,\qquad a=0.}
$$
The drop remains in motion: zero <acceleration> does not imply zero <velocity>. In a quadratic <drag force> model, $F_{\rm drag}=\rho_{\rm air}C_DAv^2/2$, giving $v_t=\sqrt{2(m-\rho_{\rm air}V)g/(\rho_{\rm air}C_DA)}$. 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>.
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