Problem source

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 source

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.