Mechanics, EXT2 EQ-Bank 31

A particle is projected from the origin with speed, \(V\), at an angle of \(\alpha\) above the horizontal. It is subject to both gravity and an air resistance proportional to its velocity, so that its horizontal and vertical components of acceleration while it is rising are given by

\(\ddot{x}=-k\dot{x}\) and \(\ddot{y} = -g-k\dot{y}\)

Show that \(\dot{x} = V\cos\,\alpha\ e^{-kt}\) and \(\dot{y} = \left( \dfrac{g}{k} + V\sin\,\alpha\right)e^{-kt}-\dfrac{g}{k} \) (2 marks)

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Show that when the particle reaches its greatest height, it has travelled a horizontal distance of
\(\dfrac{V^2\sin\,2\alpha}{2(g+Vk\,\sin\,\alpha)}\). (3 marks)

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a. \(\text{See Worked Solutions}\)

b. \(\text{See Worked Solutions}\)

Show Worked Solution

a. \(\ddot{x}=-k \dot{x} \ \Rightarrow \ \dfrac{d \dot{x}}{d t}=-k \dot{x}\)

\(\text{Solving 1st order differential equation:}\)

\(\dot{x}=A e^{-k t}\)

\(\text{Since}\ \ \dot{x}=V \cos \alpha\ \ \text{when} \ \ t=0\ \ \Rightarrow\ \ A=V \cos \alpha\)

\(\dot{x}=V \cos \alpha e^{-k t}\)

\(\ddot{y}=-g-k \dot{y}\ \Rightarrow \ \dfrac{d \dot{y}}{d t}+k \dot{y}=-g\)

\(\text{Solving 1st order differential equation:}\)

\(\dot{y}=B e^{-k t}-\dfrac{g}{k}\)

\(\text{Since} \ \ \dot{y}=V \sin \alpha \ \ \text{when} \ \ t=0:\)

\(V \sin \alpha=B e^{-k t}-\dfrac{g}{k} \ \Rightarrow \ B=\dfrac{g}{k}+V \sin \alpha\)

\(\dot{y}=\left(V \sin \alpha+\dfrac{g}{k}\right) e^{-k t}-\dfrac{g}{k}\)

b. \(\text{At max height,} \ \ \dot{y}=0\)

\(\left(V \sin \alpha+\dfrac{g}{k}\right) e^{-k t}-\dfrac{g}{k}=0 \ \Rightarrow \ e^{-k t}=\dfrac{g}{g+V k \, \sin \alpha}\ \ldots\ (1)\)

\(\text{Find horizontal distance} \ (x):\)

\(x\)	\(=\displaystyle \int_0^t V \cos \alpha\, e^{-k t}\, d t\)
	\(=-\dfrac{V \cos \alpha}{k}\big[e^{-k t}\big]_0^t\)
	\(=\dfrac{V \cos \alpha}{k}\left(1-e^{-k t}\right)\)
	\(=\dfrac{V \cos \alpha}{k}\left(1-\dfrac{g}{g+Vk\, \sin \alpha}\right)\ \ \ \text{(using (1) above)}\)
	\(=\dfrac{V \cos \alpha}{k}\left(\dfrac{g+Vk\, \sin \alpha-g}{g+Vk\, \sin \alpha}\right)\)
	\(=\dfrac{V^2 \sin \alpha\, \cos \alpha}{g+Vk\, \sin \alpha}\)
	\(=\dfrac{V^2 \sin 2 \alpha}{2(g+Vk\, \sin \alpha)}\)