Problem source

A tennis player serves from height $H$ above horizontal ground, hitting the ball downwards with speed $v$ at an angle $\alpha$ below the horizontal. The ball just clears the net of height $h$ at horizontal distance $a$ from the server and hits the ground a further horizontal distance $b$ beyond the net. Show that
$$v^2 = \frac{ g(a+b)^2(1+\tan^2\alpha)}{ 2[H-(a+b)\tan\alpha]}$$
and
$$\tan\alpha = \frac{2a+b }{ a(a+b)}H - \frac{a+b }{ ab}h \,.$$
By considering the signs of $v^2$ and $\tan\alpha$, find upper and lower bounds on $H$ for such a serve to be possible.

Solution source

We have

\begin{align*}
\rightarrow: && a &= v\cos \alpha t_{net} \\
\Rightarrow && t_{net} &= \frac{a}{v \cos \alpha} \\
\downarrow: && H-h &= v\sin \alpha  t_{net} + \frac12 g t_{net}^2 \\
&&&= a \tan \alpha + \frac12 g \frac{a^2}{v^2} \sec^2 \alpha \\
&&&= a \tan \alpha + \frac{a^2g}{2v^2}(1 + \tan^2 \alpha) \tag{*}\\
\\
\rightarrow: && a+b &= v \cos \alpha t_{ground} \\
&& t_{ground} &= \frac{a+b}{v \cos \alpha}\\
\downarrow: && H &= v\sin \alpha  t_{ground} + \frac12 g t_{ground}^2 \\
&&&= (a+b)\tan \alpha + \frac{(a+b)^2g}{2v^2}(1+\tan^2\alpha) \tag{**} \\
\\
(**): && v^2 &= \frac{g(a+b)^2(1+\tan^2\alpha)}{2[H-(a+b)\tan \alpha]} \\
(a+b)^2(*) - a^2(**): && (a+b)^2(H-h) -a^2H &= [(a+b)^2a - a^2(a+b)]\tan \alpha \\
\Rightarrow && (2ab+b^2)H - (a+b)^2h &= ab(a+b) \tan \alpha \\
\Rightarrow && \tan \alpha &= \frac{2a+b}{a(a+b)}H - \frac{a+b}{ab} h
\end{align*}

Noting that $v^2 \geq 0$ and the numerator is positive, we must have 
\begin{align*}
&& H &> (a+b)\tan \alpha \\
&&&= \frac{2a+b}{a}H - \frac{(a+b)^2}{ab} h \\
\Rightarrow && \frac{a+b}{a}H &< \frac{(a+b)^2}{ab} h \\
\Rightarrow && H &< \frac{a+b}{b} h
\end{align*}

Noting that $\tan \alpha > 0$ we must have

\begin{align*}
&& \frac{2a+b}{a(a+b)} H & > \frac{a+b}{ab} h \\
\Rightarrow && H &> \frac{(a+b)^2}{b(2a+b)}h
\end{align*}