32 problems found
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.
By substituting \(y(x)=xv(x)\) in the differential equation \[ x^{3}\frac{\mathrm{d}v}{\mathrm{d}x}+x^{2}v=\frac{1+x^{2}v^{2}}{\left(1+x^{2}\right)v}, \] or otherwise, find the solution \(v(x)\) that satisfies \(v=1\) when \(x=1\). What value does this solution approach when \(x\) becomes large?
Solution: Let \(y = xv\) then \(y' = v + xv'\) and so \(x^2y' = x^2v + x^3v'\) Our differential equation is now: \begin{align*} && x^2 y' &= \frac{1+y^2}{(1+x^2)\frac{y}{x}} \\ \Rightarrow && xy' &= \frac{(1+y^2)}{(1+x^2)y} \\ \Rightarrow && \frac{y}{1+y^2} \frac{\d y}{\d x} &= \frac{1}{x(1+x^2)} \\ \Rightarrow && \frac{y}{1+y^2} \frac{\d y}{\d x} &= \frac{1}{x} - \frac{x}{1+x^2} \\ \Rightarrow && \frac12 \ln(1+y^2) &= \ln x - \frac12 \ln(1+x^2) + C \\ \Rightarrow && \frac12 \ln (1 + y^2) &= \frac12 \ln \l \frac{x^2}{1+x^2}\r + C \\ \Rightarrow && 1+y^2 &= \frac{Dx^2}{1+x^2} \\ \Rightarrow && D &= 4 \quad \quad: (x = 1, v = 1, y = 1) \\ \Rightarrow && 1 + x^2v^2&= \frac{4x^2}{1+x^2}\\ \Rightarrow && v^2 &= \frac{3x^2-1}{x^2(1+x^2)} \\ \Rightarrow && v &= \sqrt{\frac{3x^2-1}{x^2(1+x^2)}} \\ \end{align*} As \(x \to \infty\), \(v \to 0\)