Projectiles on inclined planes with coefficient of restitution 
This is the problem I am currently attempting. So far, I've resolved the velocity parallel and perpendicular to the plane to get, perpendicular: $u \sin \theta$ upon launch and $-u\sin \theta$ on landing.
Parallel: $u \cos \theta - 2u \sin \theta \tan \alpha$.
Where do I go from here?
 A: @EDIT: SOLVED IT! :) The key part I was missing was that $e$ only acts on the component perpendicular to the slope (the y-component), i.e $u_{r,x} = v_x$ and $u_{r,x} = -e v_y$. Huge thanks to @Floris for spotting this, and all the help!!
Always start with a diagram :)

I tried solving this two ways: relative to vertical/horizontal and relative to the slope. The reflection for vertical/horizontal is horrendous, and it works out much neater to just resolve relative to the slope.
Relative to the slope of the plane


Projection
\begin{align} 
{v_x \choose v_y} &= {u_x + a_x t_1\choose u_y  + a_y t_1} \\
{s_x \choose 0} &= {u_x t_1 + \frac{1}{2}a_x t_1^2\choose u_y t_1 + \frac{1}{2}a_y t_1^2}
\end{align}
where $u_x = u\cos(\theta)$, $u_y = u\sin(\theta)$, $a_x = -mg \sin(\alpha)$, $a_y = -mg \cos(\alpha)$ and $s_y = 0$.
Reflection
$$
\vec{u_{r}} = \vec{v} - 2(\vec{v} \cdot \hat{n})\hat{n}
$$
where $\hat{n} = {0 \choose 1}$ is the unit normal vector to the slope.
$$
{u_{r,x} \choose u_{r,y}} = {v_x \choose v_y} - 2 ({v_x \choose v_y} \cdot {0 \choose 1}) {0 \choose 1}
$$
Rearrange the equation and remember that the velocity perpendicular to the plane is reduced by a factor of $e$.
$$
{u_{r,x} \choose u_{r,y}} = {v_x \choose -ev_y}
$$
Rebound
\begin{align} 
{-s_x \choose -s_y} 
&= {u_{r,x} t_2 + \frac{1}{2}a_x t_2^2\choose u_{r,y} t_2 + \frac{1}{2}a_y t_2^2} 
\\
{s_x \choose 0} &= {-v_x t_2 - \frac{1}{2}a_x t_2^2 \choose -ev_y t_2 + \frac{1}{2}a_y t_2^2}
\end{align} 
By using $s_y = 0$, we can immediately solve for $t_1$ and $t_2$
$$
t_1 = -\frac{2u_y}{a_y} \quad t_2 = \frac{2ev_y}{a_y}
$$
By plugging our solution for $t_1$ into $v_y = u_y  + a_y t_1$, we get
$$
v_y = -u_y
$$
which in hindsight is obvious, because acceleration perpendicular to the plane is constant.
Now for the fun part: setting $s_x$ during projection equal to the $s_x$ during the rebound.
$$
u_x t_1 + \frac{1}{2}a_x t_1^2 = -v_x t_2 - \frac{1}{2}a_x t_2^2
$$
Plug in for time
$$
[u_x + \frac{1}{2}a_x (-\frac{2u_y}{a_y}) ](-\frac{2u_y}{a_y}) = [-v_x - \frac{1}{2}a_x \frac{2ev_y}{a_y}] \frac{2ev_y}{a_y}
$$
Cancel $\frac{2}{a_y}$ from both sides
$$
[- u_x + u_y\frac{a_x}{a_y} ]u_y = [-v_x - ev_y \frac{a_x}{a_y}] ev_y
$$
Multiply both sides by $-1$ and factor out $u_y$ from the left and $v_y$ from the right.
$$
[\frac{u_x}{u_y} - \frac{a_x}{a_y} ]u_y^2 = [\frac{v_x}{v_y} + e\frac{a_x}{a_y}] ev_y^2
$$
Remember that $v_y = -u_y$, so we can cancel $v_y^2$ and $u_y^2$ from both sides. 
$$
\frac{u_x}{u_y} - \frac{a_x}{a_y} = e\frac{v_x}{v_y} + e^2\frac{a_x}{a_y}
$$
Now we want to plug in for $v_x/v_y$
$$
v_x = u_x + a_x t_1 = u_x -2 \frac{a_x}{a_y}u_y = (\frac{u_x}{u_y} - 2 \frac{a_x}{a_y})u_y \\ 
\therefore \frac{v_x}{v_y} = 2 \frac{a_x}{a_y} - \frac{u_x}{u_y}
$$
Plug this back in
$$
\frac{u_x}{u_y} - \frac{a_x}{a_y} = e(2 \frac{a_x}{a_y} - \frac{u_x}{u_y}) + e^2\frac{a_x}{a_y}
$$
This is a quadratic in $e$, so rearrange into an obviously quadratic form
$$
\frac{a_x}{a_y} e^2 + (2 \frac{a_x}{a_y} - \frac{u_x}{u_y})e + \frac{a_x}{a_y} - \frac{u_x}{u_y} = 0 
$$
Solve using the quadratic formula $e = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
\begin{align} 
b^2 - 4ac &= (2 \frac{a_x}{a_y} - \frac{u_x}{u_y})^2 - 4 (\frac{a_x}{a_y})(\frac{a_x}{a_y} - \frac{u_x}{u_y}) \\
&= 4 (\frac{a_x}{a_y})^2 - 4 \frac{a_x}{a_y}\frac{u_x}{u_y} + (\frac{u_x}{u_y})^2 - 4(\frac{a_x}{a_y})^2 + 4\frac{a_x}{a_y}\frac{u_x}{u_y} \\
&= (\frac{u_x}{u_y})^2
\end{align} 
$$
e = \frac{(\frac{u_x}{u_y} - 2 \frac{a_x}{a_y}) \pm \frac{u_x}{u_y}}{2\frac{a_x}{a_y}} \\
\therefore \quad
e_- = -1, \quad
e_+ = \frac{u_x}{u_y}\frac{a_y}{a_x} - 1
$$
Remember the definitions from the start: $u_x = u\cos(\theta)$, $u_y = u\sin(\theta)$, $a_x = -mg \sin(\alpha)$, $a_y = -mg \cos(\alpha)$,  so
$$
\frac{u_x}{u_y} = \cot(\theta) \quad \frac{a_x}{a_y} = \tan(\alpha)
$$
Plug this into our expression for $e_+$ and voila!
$$
e_+ = \cot(\theta) \cot(\alpha) - 1
$$
