Show that for a given velocity of projection the maximum range down an inclined plane of inclination $\alpha$ is greater than up the plane in the ratio $$\frac{1+\sin(\alpha)}{1-\sin(\alpha)}$$
Let $u$ be the given velocity of projection and $\theta$ the angle the direction of projection makes with the plane.(see picture)
I have found $x$ and $y$ components of acceleration, velocity and displacement both up and down the plane. But I could not find any light by juggling with the equations.
Taking motion up the plane
acc.= $$\begin{pmatrix}-g sin \alpha\\-gcos\alpha\end{pmatrix}$$ vel.= $$\begin{pmatrix}-gtsin\alpha +u cos\theta\\-gt cos\alpha + usin\theta\end{pmatrix}$$
Displacement=$$\begin{pmatrix}-\frac{gt^2}{2}sin\alpha +ut cos\theta\\-\frac{gt^2}{2} cos\alpha + utsin\theta\end{pmatrix}$$
Motion down plane is similar except for displacement
Displacement=$$\begin{pmatrix}-\frac{gt^2}{2}sin\alpha -ut cos\theta\\-\frac{gt^2}{2} cos\alpha - utsin\theta\end{pmatrix}$$
I have proceeded by setting $y$-component of velocity up the plane to be equal to 0 to obtain time $t$ at maximum height.Hence I found the total time of flight, T. I replaced the value of T in the $x$-component of displacement.Similar equations were drawn for motion down the plane, but I got lost on the way. I hope someone will be able to see some light now.