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, $\lim _ { \mathrm { t } \rightarrow \infty }$, of a 210 lb. skydiver $( \mathrm { mg } = 210 )$ when $\mathrm { k } = 0.006$.
A) 59.16 ft/sec
B) 187.08 ft/sec
C) There is no limiting speed.
D) 0.01 ft/sec

Question

Quizplus · Accepted Answer

To find the limiting velocity, we take the limit as $t$ approaches infinity. As $t$ gets very large, the argument of the hyperbolic tangent function approaches infinity, so the function itself approaches 1. Therefore, we have $$\lim_{t\to \infty} \sqrt{\frac{\mathrm{gk}}{\mathrm{m}}t}= \infty$$ and $$\lim_{t\to\infty} \tanh\left(\sqrt{\frac{\mathrm{gk}}{\mathrm{m}}t}\right) = 1.$$ Substituting these limits into the equation for velocity, we get $$\lim_{t\to\infty} \mathrm{v} = \sqrt{\frac{\mathrm{mg}}{\mathrm{k}}} \lim_{t\to\infty} \tanh\left(\sqrt{\frac{\mathrm{gk}}{\mathrm{m}}t}\right) = \sqrt{\frac{(210\;\mathrm{lb})(32.2\;\mathrm{ft/s}^2)}{(0.006)}}(1) \approx 187.08\;\mathrm{ft/s}.$$ Therefore, the answer is B.

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

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