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

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

A) 264.58 ft/sec
B) 0.01 ft/sec
C) There is no limiting speed.
D) 83.67 ft/sec