Answer
(a) $\lim\limits_{t \to \infty}v(t) = \sqrt{\frac{mg}{k}}$
(b) In a belly-to-earth position, the terminal velocity is $33.8~m/s$
In a feet-first position, the terminal velocity is $93.7~m/s$
Work Step by Step
(a) Consider $~~tanh(t\sqrt{\frac{g~k}{m}})$:
$tanh(t\sqrt{\frac{g~k}{m}}) = \frac{sinh(t\sqrt{\frac{g~k}{m}})}{cosh(t\sqrt{\frac{g~k}{m}})}$
$tanh( t\sqrt{\frac{g~k}{m}} ) = \frac{\frac{e^{t\sqrt{\frac{g~k}{m}}}-e^{-t\sqrt{\frac{g~k}{m}}}}{2}} {\frac{e^{t\sqrt{\frac{g~k}{m}}}+e^{-t\sqrt{\frac{g~k}{m}}}}{2}}$
$tanh( t\sqrt{\frac{g~k}{m}} ) = \frac{e^{t\sqrt{\frac{g~k}{m}}}-e^{-t\sqrt{\frac{g~k}{m}}}} {e^{t\sqrt{\frac{g~k}{m}}}+e^{-t\sqrt{\frac{g~k}{m}}}}$
When $t$ is very large, then $e^{-t\sqrt{\frac{g~k}{m}}} \approx 0$
Then:
$\lim\limits_{t \to \infty} tanh(t\sqrt{\frac{g~k}{m}}) = \frac{e^{t\sqrt{\frac{g~k}{m}}}-0}{e^{t\sqrt{\frac{g~k}{m}}}+0} = 1$
Thus:
$\lim\limits_{t \to \infty}v(t) = \lim\limits_{t \to \infty} \sqrt{\frac{mg}{k}}~tanh(t\sqrt{\frac{g~k}{m}}) = \sqrt{\frac{mg}{k}}$
(b) In a belly-to-earth position, we can find the terminal velocity:
$v = \sqrt{\frac{mg}{k}}$
$v = \sqrt{\frac{(60)(9.8)}{0.515}}$
$v = 33.8~m/s$
In a feet-first position, we can find the terminal velocity:
$v = \sqrt{\frac{mg}{k}}$
$v = \sqrt{\frac{(60)(9.8)}{0.067}}$
$v = 93.7~m/s$
