Problem source

An automated mobile dummy target for gunnery practice is moving anti-clockwise around the circumference of a large circle of radius $R$ in a horizontal plane at a constant angular speed $\omega$.  
A shell is fired from $O$, the centre of this circle, with initial speed $V$ and angle of elevation $\alpha$.  
Show that if $V^2 < gR$, then no matter what the value of $\alpha$, or what vertical plane the shell is fired in, the shell cannot hit the target.
Assume now that $V^2 > gR$ and that the shell hits the target, and let $\beta$ be the angle through which the target rotates between the time at which the shell is fired and the time of impact. Show that $\beta$ satisfies the equation
$$
g^2{{\beta}^4} - 4{{\omega}^2}{V^2}{{\beta}^2}
+4{R^2}{{\omega}^4}=0.
$$
Deduce that there are exactly two possible values of $\beta$. Let $\beta_1$ and $\beta_2$ be the possible values of $\beta$ and let $P_1$ and $P_2$ be the corresponding points of impact.
By considering the quantities $(\beta_1^2 +\beta_2^2) $ and $\beta_1^2\beta_2^2\,$, or otherwise, show that the linear distance between $P_1$ and $P_2$ is
\[
2R \sin\Big( \frac\omega g \sqrt{V^2-Rg}\Big) 
\;.
\]

Solution source

\begin{align*}
&& 0 &= V\sin \alpha t-\frac12 gt^2 \\
\Rightarrow && t &= \frac{2V \sin \alpha}{g} \\
&& R &= V \cos \alpha \, t \\
&&&= \frac{2V^2 \sin \alpha \cos \alpha}{g} \\
&&&= \frac{V^2 \sin 2 \alpha}{g}
\end{align*}

Therefore the max distance is $\frac{V^2}{g}$, therefore we cannot hit the target if $R > \frac{V^2}{g} \Rightarrow gR > V^2$.

We have $\beta = \omega t \Rightarrow t = \frac{\beta}{\omega}$

\begin{align*}
&& \sin \alpha &= \frac{gt}{2V} \\
&& \cos \alpha &= \frac{R}{Vt} \\
\Rightarrow && 1 &= \left (\frac{gt}{2V} \right)^2 + \left ( \frac{R}{Vt}  \right)^2 \\
&&&=  \left (\frac{g\beta}{2V \omega} \right)^2 + \left ( \frac{R\omega}{V\beta}  \right)^2 \\
&&&= \frac{g^2 \beta^2}{4 V^2 \omega^2} + \frac{R^2 \omega^2}{V^2 \beta ^2} \\
\Rightarrow && 4V^2 \omega^2 \beta^2 &= g^2 \beta^4 + 4R^2 \omega^4 \\
\Rightarrow && 0 &= g^2 \beta^4 - 4\omega^2 V^2 \beta^2+4R^2\omega^4
\end{align*}

This (quadratic) equation in terms of $\beta^2$ has two solution if $\Delta = 16\omega^4V^4-16g^2R^2\omega^4 =16\omega^4(V^4-g^2R^2) > 0$ since $V^2 > gR$.

Since $\beta > 0$ there are exactly two solutions, once we have values for $\beta$.

First notice, 

\begin{align*}
&& \beta_1^2 + \beta_2^2 &= \frac{4\omega^2V^2}{g^2} \\
&& \beta_1^2\beta_2^2 &= \frac{4R^2\omega^4}{g^2}
\end{align*} 

Then notice the positions of $P_1$ and $P_2$ are $(R\cos \beta_1 , R\sin \beta_1)$ and $(R\cos \beta_2, R\sin \beta_2)$. 

\begin{align*}
&& d^2 &= R^2\left ( \cos \beta_1 - \cos \beta_2 \right)^2 + R^2 \left ( \sin \beta_1 - \sin \beta_2 \right)^2 \\
&&&= 2R^2 - 2R^2(\cos \beta_1 \cos \beta_2 + \sin \beta_1 \sin \beta_2) \\
&&&= 2R^2-2R^2\cos(\beta_1 - \beta_2) \\
&&&= 2R^2 \left (1-\cos(\sqrt{(\beta_1-\beta_2)^2} \right ) \\
&&&= 2R^2 \left (1 - \cos\left ( \sqrt{\frac{4\omega^2 V^2}{g^2} - \frac{4R\omega^2}{g}} \right) \right) \\
&&&= 2R^2 \left (1 - \cos\left (\frac{2\omega}{g} \sqrt{V^2 - Rg} \right) \right) \\
&&&= 4 R^2 \sin^2 \left (\frac{\omega}{g} \sqrt{V^2 - Rg} \right)
\end{align*}

which gives us the required result.