For downward speed v and quadratic drag D = k v^2, ignore buoyancy and hold m, g and k constant. The terminal balance mg = k vT^2 gives vT = sqrt(mg/k). Air density, shape and area are included in k.
Worked example
Let m = 2 kg, g = 10 m/s^2 and k = 0.2 kg/m. Then vT = sqrt(20/0.2) = 10 m/s. This is a model value, not a safe landing speed.
A schematic speed-time curve rises from zero and levels toward a dashed terminal speed. Its slope gets smaller: acceleration decreases as drag approaches weight. The ideal limit is approached, not reached at a finite time.
