You can't answer that question exactly because the terminal velocity depends on many variables that change from situation to situation. Terminal velocity comes about when the frictional forces of the air are equal to the force of gravity. The frictional force of air depends on air pressure and density, shape of the falling body, etc. For example the terminal velocity of a sheet of paper or a balloon is much smaller than the terminal of a BB weighing the same amount.
However, you can make an estimate for a human body. It is said, although I've never tried it myself, that the terminal velocity of an ordinary person falling in normal atmosphere at low altitudes is around V
0 = 200 kilometers per hour ( ~ 120 mph). Using the standard equations of motion and assuming that the air resistance force is proportional to the velocity squared then you can solve for the velocity and distance. There are two parameters in the solution in addition to V
0: the characteristic time,
T
0 = V
0/g = 5.6 sec, and the characteristic distance, X
0 = V
0T
0 = 315 m.
The full solution is V = V
0 tanh(T/T
0) and X = X
0 log( cosh(T/T
0) ) .
Notice that V only approaches V
0 asymptotically, it never really gets there.
For T = T
0, you get V is about 3/4 the terminal velocity and you will have
fallen about 136 m.
LeeH
(published on 03/27/2010)
