A particle is projected from a point on a plane that is inclined at an angle~\(\phi\) to the horizontal. The position of the particle at time \(t\) after it is projected is \((x,y)\), where \((0,0)\) is the point of projection, \(x\) measures distance up the line of greatest slope and \(y\) measures perpendicular distance from the plane. Initially, the velocity of the particle is given by \((\dot x, \dot y) = (V\cos\theta, V\sin\theta)\), where \(V>0\) and \(\phi+\theta<\pi/2\,\). Write down expressions for \(x\) and \(y\). The particle bounces on the plane and returns along the same path to the point of projection. Show that \[2\tan\phi\tan\theta =1\] and that \[ R= \frac{V^2\cos^2\theta}{2g\sin\phi}\,, \] where \(R\) is the range along the plane. Show further that \[ \frac{2V^2}{gR} = 3\sin\phi + {\rm cosec}\,\phi \] and deduce that the largest possible value of \(R\) is \(V^2/ (\sqrt{3}\,g)\,\).