A parachuter and his parachute together weigh $192 {\rm lb}$. At the instant the parachute opens, he is falling at 150 feet per second (ft/s). Suppose the air resistance exerts a force of 200 lb when the parachuter’s speed is 20 ft/s.

Solution (a) In the U.S. customary system of units, length is measured in feet, weight in pounds, time in seconds, and $g\approx 32\;\rm{ft}/\rm{s}^{2}$. The mass $m$ of the parachuter and his parachute obeys $mg=\hbox{weight}=192\;\rm{lb}$.

Since the air resistance is $200\;\rm{lb}$ when the speed is $20\;\rm{ft}/\rm{s}$, $$\begin{eqnarray*} 200& =&k\cdot 20\qquad\color{#0066A7}{{F=kv}; \hbox{force is proportional to speed.}} \\ k& =&10 \end{eqnarray*}$$

From (3), the velocity of the parachuter at time $t$ after the parachute opens is $$\begin{eqnarray*} v(t) &=&-\frac{mg}{k}+\left( v_{0}+\frac{mg}{k}\right) e^{{{-kt/m}}} \\ v(t) &=&-\frac{192}{10}+\left( -150+\frac{192}{10}\right) e^{{{-10t/6}}}\qquad\color{#0066A7}{{v}_{0}=-150; mg=192; g=32; k=10.} \\ v(t) &=&-19.2-130.8e^{{{-5t/3}}} \end{eqnarray*}$$

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After $3$ seconds, the parachuter’s velocity is $$\begin{equation*} v(3)=-19.2-130.8e^{-5}\approx -20.081\;\rm{ft}/\!\rm{s} (13.692 \;\rm{mi}/\!\rm{h}) \end{equation*}$$

(b) The limiting velocity is $\lim\limits_{t\rightarrow \infty }v(t)=-\dfrac{mg}{k}=-19.2\;\rm{ft}/\rm{s} (13.091\;\rm{mi}/\rm{h} )$.