Textbook Problem

A model for the velocity of a falling object after time t is

$v(t)=\sqrt{\frac{mg}{k}\tanh \left(t\sqrt{\frac{gk}{m}}\right)}$

where m is the mass of the object, g = 9.8 m/s² is the acceleration due to gravity, k: is a constant, t is measured in seconds, and v in m/s.

(a) Calculate the terminal velocity of the object, that is, lim_t→∞ v(t).

(b) If a person falls from a building, the value of the constant k depends on his or her position. For a “belly-to-earth” position, k = 0.515 kg/s, but for a “feet-fist” position, k = 0.067 kg/s. If a 60-kg person falls in belly-to-earth position, what is the terminal velocity? What about feet-first?

Source: L. Long et at, “How Terminals Terminal Velocity?” American Mathematical Monthly 113 (2006): 752–55.