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Solve the Problem mm Falling from Rest Under the Action of Gravity Is Given

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Solve the problem.
-The velocity of a body of mass mm falling from rest under the action of gravity is given by the equation v=mgktanh(gkmt) \mathrm { v } = \sqrt { \frac { \mathrm { mg } } { \mathrm { k } } } \tanh \left( \sqrt { \frac { \mathrm { gk } } { \mathrm { m } } } \mathrm { t } \right) , where k\mathrm { k } is a constant that depends on the body's aerodynamic properties and the density of the air, gg is the gravitational constant, and tt is the number of seconds into the fall. Find the limiting velocity, limtv\lim _ { \mathrm { t } \rightarrow } \mathrm { v } , of a 420lb420 \mathrm { lb } . skydiver ( mg=420\mathrm { mg } = 420 ) when k=0.006\mathrm { k } = 0.006 .


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