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Mechanics, EXT2 EQ-Bank 36

An experimental rocket is at a height of 5000 m, ascending at a speed of \(50\sqrt{2}\) m s\(^{-1}\) at an angle of 45° to the horizontal, when its engine stops. The rocket is then subject to gravity and to air resistance proportional to its velocity. Take \(g\) = 10 m s\(^{-2}\).

The velocity vector of the rocket, \(t\) seconds after the engine stops, is

\(\mathbf{v}(t) = 50e^{-0.2t}\,\mathbf{i} + (100e^{-0.2t}-50)\mathbf{j}.\)   (Do NOT prove this.)
 

  1. Show that the rocket reaches its greatest height when  \(t =5\ln 2\)  seconds, and calculate its greatest height.   (3 marks)

  2. The pilot can only operate the ejection seat while the rocket is descending at an angle between 45° and 60° to the horizontal. Find the earliest and latest times at which the pilot can eject.   (3 marks)

  3. As the rocket continues to fall, its speed approaches a limiting value. Find this terminal speed, justifying your answer.   (1 mark)

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a.    \(\text{See Worked Solutions}\)

b.    \(\text{Pilot can eject between 5.5 and 6.6 seconds.}\)

c.    \(\text{Terminal speed}=50 \ \text{ms}^{-1}\)

Show Worked Solution

a.    \(\mathbf{v}(t)=50 e^{-0.2 t}\,\mathbf{i}+\left(100 e^{-0.2 t}-50\right)\mathbf{j}\)

\(\text{Max height occurs when} \ \ \dot{y}=0:\)

\(100 e^{-0.2 t}-50\) \(=0\)
\(e^{-0.2 t}\) \(=\dfrac{1}{2}\)
\(-0.2 t\) \(=-\ln 2\)
\(t\) \(=5 \ln 2\)

 
\(\text{Find} \ y  \ \text{when}\ \  t=5 \ln 2:\)

\(y(t)=\displaystyle \int 100 e^{-0.2 t}-50\, d t=-500 e^{-0.2 t}-50 t+c\)

\(\text{When} \ \ t=0, y=5000:\)

\(5000=-500 e^{\circ}+c \ \Rightarrow \ c=5500\)

\(y=5500-500 e^{-0.2 t}-50 t\)
 

\(\text{At} \ \ t=5\ln 2:\)

\(y=5500-500 e^{-\ln 2}-50 \times 5 \ln 2=5076.71 \ldots=5077 \ \text{m}\).
 

b.    \(\text {On descent,} \ \ \dot{y}<0.\)

\(\text{Let} \ \ \theta=\text{angle below the horizontal}\)

\(\tan \theta=\dfrac{\abs{\dot{y}}}{\dot{x}}\)

\(\text{Let}\ \  a=e^{-0.2 t}\)

\(\tan \theta=\dfrac{50-100 a}{50 a}=\dfrac{1}{a}-2 \ \Rightarrow \ \theta=\tan ^{-1}\left(\dfrac{1}{a}-2\right)\)
 

\(\text{When} \ 45^{\circ} \ \text {is reached:}\)

\(\dfrac{1}{a}-2=1 \ \Rightarrow \ a=3\)

\(e^{-0.2 t}=\dfrac{1}{3} \ \Rightarrow \ t=\dfrac{\ln 3}{0.2} \approx 5.5 \ \text{s  (1 d.p.)}\)
 

\(\text{When} \ 60^{\circ} \ \text {is reached:}\)

\(\dfrac{1}{a}-2=\sqrt{3} \ \Rightarrow \ a=\dfrac{1}{2+\sqrt{3}}=2-\sqrt{3}\)

\(e^{-0.2 t}\) \(=2-\sqrt{3}\)
\(-0.2 t\) \(=\ln (2-\sqrt{3})\)
\(t\) \(=-5\ln (2-\sqrt{3}) \approx 6.6 \ \text{s  (1 d.p.)}\)

 
\(\therefore \ \text{Pilot can eject between 5.5 and 6.6 seconds.}\)
 

c.    \(\text{As} \ \ t \rightarrow \infty:\)

\(e^{-0.2 t} \rightarrow 0\ \ \Rightarrow\ \ \dot{x} \rightarrow 0, \ \ \dot{y} \rightarrow -50\)

\(\therefore \ \text{Terminal speed}=\sqrt{0^2+50^2}=50 \ \text{ms}^{-1}\)