Show that, if a particle is projected at an angle \(\alpha\) above the horizontal with speed \(u\), it will reach height \(h\) at a horizontal distance \(s\) from the point of projection where
\[h = s\tan\alpha - \frac{gs^2}{2u^2\cos^2\alpha}\,.\]
The remainder of this question uses axes with the \(x\)- and \(y\)-axes horizontal and the \(z\)-axis vertically upwards. The ground is a sloping plane with equation \(z = y\tan\theta\) and a road runs along the \(x\)-axis. A cannon, which may have any angle of inclination and be pointed in any direction, fires projectiles from ground level with speed \(u\). Initially, the cannon is placed at the origin.
Let a point \(P\) on the plane have coordinates \((x,\, y,\, y\tan\theta)\). Show that the condition for it to be possible for a projectile from the cannon to land at point \(P\) is
\[x^2 + \left(y + \frac{u^2\tan\theta}{g}\right)^2 \leqslant \frac{u^4\sec^2\theta}{g^2}\,.\]
Show that the furthest point directly up the plane that can be reached by a projectile from the cannon is a distance
\[\frac{u^2}{g(1+\sin\theta)}\]
from the cannon.
How far from the cannon is the furthest point directly down the plane that can be reached by a projectile from it?
Find the length of road which can be reached by projectiles from the cannon.
The cannon is now moved to a point on the plane vertically above the \(y\)-axis, and a distance \(r\) from the road. Find the value of \(r\) which maximises the length of road which can be reached by projectiles from the cannon. What is this maximum length?