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.

How fast is he falling \(3\) seconds after the parachute opens?

What is the parachuter’s limiting velocity?

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 \(mg=\hbox{weight}=192\;\rm{lb}\).

Remember that with falling objects, the positive direction is up, so \(v_0 = v(0) =-150\;\rm{ft}/\!\rm{s}\).

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