Problem source

A fielder, who is perfectly placed to catch a ball struck by the batsman in a game of cricket, watches the ball in flight. 
Assuming that the ball is struck at the fielder's eye level and is caught just in front of her eye,
show that $\frac{ {\rm d}}{{\rm d t}} (\tan\theta ) $ is constant, where $\theta$ is the angle between the horizontal and the fielder's line of sight.
In order to catch the next  ball, which is also struck towards her but at a different velocity, the fielder runs at constant speed $v$ towards the batsman. Assuming that the ground is horizontal, show that the fielder should choose $v$ so that $\frac{ {\rm d}}{{\rm d t}} (\tan\theta ) $
remains constant.

Solution source

Set up a coordinate frame such that the position of the catch is the origin and the time of the catch is $t = 0$ . We must have then that the trajectory of the ball is $\mathbf{s} =\mathbf{u} t  + \frac12 \mathbf{g} t^2 = \binom{u_x t}{u_y t - \frac12 gt^2}$. 

We must then have: 

\begin{align*}
&& \tan \theta &= \frac{u_y t - \frac12 gt^2}{u_x t} \\
&&&=  \frac{u_y}{u_x} - \frac{g}{2u_x} t \\
\Rightarrow && \frac{\d}{\d \theta} \left ( \tan \theta \right) &= 0 - \frac{g}{2 u_x}
\end{align*}

which is clearly constant.

Set coordinates so $y$-axis starts from eye-level and $t = 0$ the first time the ball reaches that level. (Or move the trajectory backwards if that's not the case).

Then the ball has trajectory $\binom{u_xt}{u_yt - \frac12 gt^2}$. The ball reaches eye level a second time when $t = \frac{2u_y}{g}$, ie at a point $\frac{2u_xu_y}{g}$.

The fielder therefore needs to have position $f + (u_x-\frac{g}{2u_y}f)t$ at all times.

Therefore

\begin{align*}
&& \tan \theta &= \frac{u_y t - \frac12 gt^2}{f + (u_x-\frac{g}{2u_y}f)t - u_x t} \\
&&&= \frac{u_y t - \frac12 gt^2}{f(1-\frac{g}{2u_y}t)} \\
&&&= \frac{u_yt ( 1- \frac{g}{2u_y}t)}{f( 1- \frac{g}{2u_y}t)} \\
&&&= u_y t \\
\Rightarrow && \frac{\d}{\d \theta} \left ( \tan \theta \right) &= u_y
\end{align*}

Ie $ \frac{\d}{\d \theta} \left ( \tan \theta \right) $ is constant as required.