2 problems found
A particle of mass \(m\) is projected with velocity \(\+ u\). It is acted upon by the force \(m\+g\) due to gravity and by a resistive force \(-mk \+v\), where \(\+v\) is its velocity and \(k\) is a positive constant. Given that, at time \(t\) after projection, its position \(\+r\) relative to the point of projection is given by \[ \+r = \frac{kt -1 +\.e^{-kt}} {k^2} \, \+g + \frac{ 1-\.e^{-kt}}{k} \, \+u \,, \] find an expression for \(\+v\) in terms of \(k\), \(t\), \(\+g\) and \(\+u\). Verify that the equation of motion and the initial conditions are satisfied. Let \(\+u = u\cos\alpha \, \+i + u \sin\alpha \, \+j\) and $\+g = -g\, \+j\(, where \)0<\alpha<90^\circ\(, and let \)T$ be the time after projection at which \(\+r \,.\, \+j =0\). Show that \[ uk \sin\alpha = \left(\frac{kT}{1-\.e^{-kT}} -1\right)g\,. \] Let \(\beta\) be the acute angle between \(\+v\) and \(\+i\) at time \(T\). Show that \[ \tan\beta = \frac{(\.e^{kT}-1)g}{uk\cos\alpha}-\tan\alpha \,. \] Show further that \(\tan\beta >\tan\alpha\) (you may assume that \(\sinh kT >kT\)) and deduce that~\(\beta >\alpha\).
A goalkeeper stands on the goal-line and kicks the football directly into the wind, at an angle \(\alpha\) to the horizontal. The ball has mass \(m\) and is kicked with velocity \(\mathbf{v}_{0}.\) The wind blows horizontally with constant velocity \(\mathbf{w}\) and the air resistance on the ball is \(mk\) times its velocity relative to the wind velocity, where \(k\) is a positive constant. Show that the equation of motion of the ball can be written in the form \[ \frac{\mathrm{d}\mathbf{v}}{\mathrm{d}t}+k\mathbf{v}=\mathbf{g}+k\mathbf{w}, \] where \(\mathbf{v}\) is the ball's velocity relative to the ground, and \(\mathbf{g}\) is the acceleration due to gravity. By writing down horizontal and vertical equations of motion for the ball, or otherwise, find its position at time \(t\) after it was kicked. On the assumption that the goalkeeper moves out of the way, show that if \(\tan\alpha=\left|\mathbf{g}\right|/(k\left|\mathbf{w}\right|),\) then the goalkeeper scores an own goal.
Solution: Applying \(\mathbf{F} = m\mathbf{a} = m \frac{\d \mathbf{v}}{dt}\) we have: \begin{align*} && m \frac{\d \mathbf{v}}{d t} &= m\mathbf{g} - mk(\mathbf{v} - \mathbf{w}) \\ \Rightarrow && \frac{\d \mathbf{v}}{d t} +k \mathbf{v} &= \mathbf{g} + k \mathbf{w} \\ \\ \Rightarrow && e^{k t} \l \frac{\d \mathbf{v}}{d t} +k \mathbf{v} \r &= e^{kt} ( \mathbf{g} + k \mathbf{w}) \\ \Rightarrow && \frac{\d}{\d t} \l e^{kt} \mathbf{v} \r &= e^{kt}( \mathbf{g} + k \mathbf{w}) \\ \Rightarrow && e^{kt} \mathbf{v} &= \frac{1}ke^{kt}( \mathbf{g} + k \mathbf{w}) + c \\ \Rightarrow && \mathbf{v}_0 &= \frac{1}{k} ( \mathbf{g} + k \mathbf{w})+c \\ \Rightarrow && \mathbf{v} &= e^{-kt} \l \mathbf{v_0} - \frac{1}{k}\mathbf{g} - \mathbf{w} \r + \frac{1}{k} \mathbf{g} + \mathbf{w} \\ \Rightarrow && \mathbf{x} &= -\frac{1}{k}e^{-kt} \l \mathbf{v_0} - \frac{1}{k}\mathbf{g} - \mathbf{w} \r + \frac{1}{k} \mathbf{g}t + \mathbf{w}t+C \\ \Rightarrow && \mathbf{0} &= -\frac{1}{k} \l \mathbf{v_0} - \frac{1}{k}\mathbf{g} - \mathbf{w} \r + C \\ \Rightarrow && \mathbf{x} &= \frac1{k}\l 1- e^{-kt} \r\l \mathbf{v_0} - \frac{1}{k}\mathbf{g} - \mathbf{w} \r + \frac{1}{k} \mathbf{g}t + \mathbf{w}t \end{align*} Position at time \(t\) is: \begin{align*} && x_x &= \frac1{k} ( 1-e^{-kt})(u_x - w)+wt \\ && x_y &= \frac1{k} ( 1-e^{-kt})(u_x \frac{g}{kw} - \frac{g}{k})+\frac{1}{k}gt \\ &&&= \frac{g}{kw} \left ( ( 1-e^{-kt})(u_x - w)+wt \right) \\ &&&= \frac{g}{kw} x_x \end{align*} Therefore if \(x_x\) is ever \(0\) then \(x_y\) will also be zero. But the ball must eventually hit the ground, and when it does, it will be in the process of scoring an own goal.