Problem source

A particle is projected over level ground with a speed  $u$ at 
an angle $\theta$ above the horizontal. 
Derive
 an expression for the greatest height of the particle in terms of $u$, $\theta$ and $g$. 

A  particle is  projected from the floor of a horizontal tunnel of height ${9\over 10}d$. 
Point $P$
 is  ${1\over 2}d$ metres vertically and $d$ metres horizontally along the tunnel
 from the point of 
projection. The particle passes through point $P$ and lands inside the tunnel
without hitting the roof. Show that
\[
\arctan \textstyle {3 \over 5} < \theta < \arctan \, 3 \;.
\]

Solution source

\begin{align*}
&& v^2 &= u^2 + 2as \\
(\uparrow): && 0 &= (u \sin \theta)^2 - 2gh \\
\Rightarrow && h &= \frac{u^2 \sin^2 \theta}{2g}
\end{align*}

To avoid hitting the ceiling \begin{align*}
&& \frac9{10}d &>\frac{u^2 \sin^2 \theta}{2g} \\
\Rightarrow && \frac{gd}{u^2}&>\frac{5\sin^2 \theta}{9} \\
\end{align*}

In order to pass through $P$ we need 
\begin{align*}
&& d &= u \cos \theta t \\
&& \frac12 d &= u \sin \theta t - \frac12 g t^2 \\
\Rightarrow && \frac12 &=  \tan \theta - \frac12 \frac{g d}{u^2} \sec^2 \theta \\
&&&<\tan \theta - \frac12 \frac{5 \sin^2 \theta}{9\cos^2 \theta} \\
\Rightarrow && 0 & >5 \tan^2 \theta -18\tan \theta +9 \\
&&&= (5\tan \theta - 3)(\tan \theta - 3) \\
\\
\Rightarrow && \tan \theta &\in \left ( \frac35, 3\right) \\
&& \theta &= \left (\arctan \tfrac35, \arctan 3 \right)
\end{align*}