A model for the velocity of a falling object after time $ t $ is
$ v(t) = \sqrt {\frac {mg}{k}} \tanh (t \sqrt { \frac {gk}{m}}) $
where $ m $ is the mass of the object, $ g = 9.8 m/s^2 $ 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 objects, that is, $ \lim_{t \to ^x} 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-first" 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?