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Using the usual suspect for a first estimate, we recall
the parachutist falling at 120 mph (54 m/s)
We say he weighs 180 lb (803N),
and we say weight = kAv^2 where air drag balances weight
so 803 = k * 1.11 * 54 * 54 and k = 0.25 in this system of units
for a jumper whose cross section A is roughly 2ft x 6 ft (1.11 m^2)
Blithely assuming that Re is unimportant, we scale to a 9mm
bullet, taking the weight as 115 grains (0.073N) and the side area
as 9 x 27 mm^2 = 2.43E-4 m^2
We jump immediately to a terminal velocity given by
0.073 = 0.25 * 2.43E-4 * v^2
V^2 = 1.2E3
V = 35 m/s (78 mph)
This is probably rather too fast for a 9 mm bullet
dropped on its side.