Mathematics Extension 2

A projectile is launched from $O$ at $V\,\mathrm{m/s}$ so that it will hit the point $A(a,b)$ on the slope. Air resistance is neglected and the acceleration due to gravity is $g\,\mathrm{m/s^2}$. Given that
$y=-\frac{gx^2}{2V^2}(1+\tan^2\alpha)+x\tan\alpha,$
and that $\alpha_1<\alpha_2$, show that
$\tan\alpha_1=\frac{V^2-\sqrt{V^4-g^2a^2-2V^2gb}}{ga}$
and
$\tan\alpha_2=\frac{V^2+\sqrt{V^4-g^2a^2-2V^2gb}}{ga}.$

By showing that
$\tan(\alpha_1+\alpha_2)=\tan(90^\circ+\theta),$
show that
$\angle COY=\angle BOA.$

A particle is projected from the origin with initial velocity $u\,\mathrm{m/s}$ at an angle $\theta$ to the horizontal. The particle lands at $x=R$ on the $x$-axis. The acceleration vector is
$\mathbf a=\begin{pmatrix}0\\-g\end{pmatrix},$
where $g$ is the acceleration due to gravity. (Do NOT prove this.)

(i) Show that the position vector $\mathbf r(t)$ of the particle is given by
$\mathbf r(t)=\begin{pmatrix}ut\cos\theta\\ut\sin\theta-\frac12gt^2\end{pmatrix}.$

Using the projectile model from part (i), show that the Cartesian equation of the path of flight is
$y=\frac{-gx^2}{2u^2}\left(\tan^2\theta-\frac{2u^2}{gx}\tan\theta+1\right).$

Given $u^2>gR$, prove that there are two distinct values of $\theta$ for which the particle will land at $x=R$.