What is the optimal angle when throwing a stone on a slope? The question what-is-the-optimum-angle-of-projection-when-throwing-a-stone-off-a-cliff was asked and answered a while back.  This one has a much cleaner answer.  Now you are on a uniform slope, a line through the origin that is not horizontal and you want to throw a stone as far as possible.  As a function of the angle of the slope, what angle should you throw at?  Usual assumptions:  uniform gravity, no friction.
 A: I am surprised no one seems to have answered this.
I get $$\pi/4 + \alpha/2$$.
If the line is given by $y = x \tan \alpha$, with $\alpha$ acute and we throw from the origin at an angle $\theta$ from the x-axis, at velocity $1$, then we have that the projectile satisfies, assuming gravity $g=2$ (in appropriate units)
$\displaystyle y = t\sin\theta - t^2$, $\displaystyle x = t\cos \theta$
The time at which it intersects the line again is given by
$\displaystyle t\sin\theta - t^2 = t \tan\alpha \cos \theta$ and so
$\displaystyle t = \sin \theta - \tan \alpha \cos \theta$
It is enough to maximize the horizontal distance travelled by the projectile, which is given by
$\displaystyle \cos \theta (\sin \theta - \tan \alpha \cos \theta)$
With little manipulation, we need to maximize
$\displaystyle  \sin(2\theta - \alpha)$
which gives the result.
A: When the solution is so neat, I feel like it should have a high-level explanation "from the book" that makes it immediately obvious. Sadly, I was not able to find one, but I'm posting what I did get in hopes that it might inspire someone else.
Imagine firing projectiles at all angles from the origin. Assume unit speed and gravity $g = 2$, as in Moron's answer. Consider a frame of reference which starts at the origin and falls freely. There is no gravity in this frame of reference, so all projectiles expand radially outward forming a cone,
$$x^2 + y^2 = t^{2}.$$
However, the ground is no longer given by $y = x \tan\alpha$ but rather by
$$y = x\tan\alpha + t^{2},$$
which is a parabolic cylinder. The projectiles hit the ground at the points where the cone and the parabolic cylinder intersect. Subtracting the two equations, we find that this intersection satisfies
$$\left(x^2 + x\tan\alpha\right) + \left(y^2 - y\right) = 0,$$
i.e. on the $xy$ plane it projects to a circle. On physical grounds, one can see that the circle passes through the origin, and is tangent to the line $y = x\tan\alpha$ there. The projectile that reaches the farthest corresponds to the point with the largest $x$-coordinate, and the angle of the projectile is simply the angle of the line joining that point to the origin (because projectiles just move radially outward in this frame of reference). I encourage you to draw a little diagram and see that this line bisects the angle between the vertical, at $\pi/2$, and the ground, at $\alpha$.
