A particle \(P\) is projected in the \(x\)-\(y\) plane, where the \(y\)-axis is vertical and the \(x\)-axis is horizontal. The particle is projected with speed \(V\) from the origin at an angle of \(45 ^\circ\) above the positive \(x\)-axis. Determine the equation of the trajectory of \(P\). The point of projection (the origin) is on the floor of a barn. The roof of the barn is given by the equation \(y= x \tan \alpha +b\,\), where \(b>0\) and \(\alpha\) is an acute angle. Show that, if the particle just touches the roof, then \(V(-1+ \tan\alpha) =-2 \sqrt{bg}\); you should justify the choice of the negative root. If this condition is satisfied, find, in terms of \(\alpha\), \(V\) and \(g\), the time after projection at which touching takes place. A particle \(Q\) can slide along a smooth rail fixed, in the \(x\)-\(y\) plane, to the under-side of the roof. It is projected from the point \((0,b)\) with speed \(U\) at the same time as \(P\) is projected from the origin. Given that the particles just touch in the course of their motions, show that \[ 2 \sqrt 2 \, U \cos \alpha = V \big(2 + \sin\alpha\cos\alpha -\sin^2\alpha) . \]