Problem source

A shot-putter projects a shot at an angle $\theta$ above the horizontal, releasing it at height $h$ above the level ground, with speed $v$.
Show that the distance $R$ travelled horizontally by the shot from its point of release until it strikes the ground is given by 
\[
R=\frac{v^{2}}{2g}\sin2\theta\left(1+\sqrt{1+\frac{2hg}{v^{2}\sin^{2}\theta}}\right).
\]
The shot-putter's style is such that currently $\theta=45^{\circ}$.
Determine (with justification) whether a small decrease in $\theta$ will increase $R$. 
[Air resistance may be neglected.]

Solution source

Notice that $u_x = v \cos \theta, u_y = v \sin \theta$.

We are interested in the time taken for the shot to hit the ground. $-h = u_y t -\frac12 g t^2$ since our distance will be $v \cos \theta \cdot t$.

Solving this quadratic for $t$ we obtain:

\begin{align*}
&& 0 &= h + v \sin \theta  \cdot t - \frac12 g \cdot t^2 \\
\Rightarrow && t_\pm &= \frac{-v \sin \theta \pm \sqrt{v^2 \sin^2 \theta+2hg}}{-g} \\
\Rightarrow && t_- &= \frac{v \sin \theta+v\sin \theta \sqrt{1 + \frac{2hg}{v^2 \sin^2 \theta}}}{g} \\
\Rightarrow && v \cos \theta t_{-} &= \frac{v^2}{g} \cos \theta \sin \theta \left( 1 + \sqrt{1 + \frac{2hg}{v^2 \sin^2 \theta}} \right) \\
&&&= \frac{v^2}{2g} \sin 2 \theta \left( 1 + \sqrt{1 + \frac{2hg}{v^2 \sin^2 \theta}} \right)
\end{align*}

Differentiating $R$ wrt to $\theta$ at $\frac{\pi}{4}$ we obtain:

\begin{align*}
\frac{\d R}{\d \theta} &= \frac{v^2}{2g} \left (2 \cos 2 \theta +  2 \cos 2 \theta \sqrt{1 + \frac{2hg}{v^2 \sin^2 \theta}} + \sin 2\theta \left ( 1 + \frac{2hg}{v^2 \sin^2 \theta}\right)^{-\frac12} \frac12 \frac{2hg} {v^2}(-2) \frac{\cos \theta}{\sin^3 \theta}\right) \\
\frac{\d R}{\d \theta} \biggr \rvert_{\theta = \frac{\pi}{4}} &=\frac{v^2}{2g}\left(0+0- 4\left ( 1 + \frac{4hg}{v^2 }\right)^{-\frac12} \frac{hg} {v^2} \right) \\
&< 0
\end{align*}

Therefore, since $R$ is locally decreasing in $\theta$ he should reduce his angle of projection slightly.