A particle is projected with speed 49 m s$^{-1}$ at an angle of elevation $\theta$ from a point $O$ on a horizontal plane, and moves freely under gravity. The horizontal and upward vertical displacements of the particle from $O$ at time $t$ seconds after projection are $x$ m and $y$ m respectively.

\begin{enumerate}[label=(\roman*)]
\item Express $x$ and $y$ in terms of $\theta$ and $t$, and hence show that
$$y = x \tan \theta - \frac{x^2(1 + \tan^2 \theta)}{490}.$$ [4]
\end{enumerate}

\includegraphics{figure_8}

The particle passes through the point where $x = 70$ and $y = 30$. The two possible values of $\theta$ are $\theta_1$ and $\theta_2$, and the corresponding points where the particle returns to the plane are $A_1$ and $A_2$ respectively (see diagram).

\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item Find $\theta_1$ and $\theta_2$. [4]
\item Calculate the distance between $A_1$ and $A_2$. [5]
\end{enumerate}

\hfill \mbox{\textit{OCR M2  Q8 [13]}}