Projectile on inclined plane, angle for maximum range A Ball is thrown up an inclined plane (incline angle $\alpha$) with $\vec{v_0}$ being at an angle of $\theta$ with the inclined plane. I added a picture to show the situation 
I already derived the angle of $\theta$ which maximises the distance on the inclined plane which turns out to be $\theta = \frac{\pi}{4} - \frac{\alpha}{2}$. This also seems to be correct when I compare it to other results on the internet.
However, is there an intuitive way for why this has to be the case? 
 A: There is no intuitive way to demonstrate that it has to be the case. But intuition usually helps in accepting that an answer is correct if you compare it to well-known situations, or extreme conditions.
In your case it would be taking the cases where $\alpha=0, \pi/2, \pi$.
In all these "extreme" conditions your equation seems to work; but it definitely doesn't mean that it has to be correct. 
A: Galileo, in his book Two new sciences, stated that “for elevations which exceed or fall short of 45° by equal amounts, the ranges are equal”.

For angle of projections $\theta$ and $(90^{\circ} - \theta + \alpha)$, the ranges on inclined plane are same. Intuitively, Range on inclined plane will be maximum, when
$$\theta=45^{\circ} \pm \dfrac{\alpha}{2}$$
$$R_{max}=\dfrac{u^2}{g(1 \pm \sin \alpha)}$$
See also this question which might give you some more intuition, Why doesn't the optimal angle (for maximum range) on an inclined plane equal 45 degrees?

A: In the case where $\alpha = 0$ the $\theta$ that maximizes the range is exactly halfway between $0$ and $\frac{\pi}{2}$ , i.e. $\frac{\pi}{4}$.
Intuitively, when $\alpha \neq 0$ the $\theta$ that will maximize the range will be exactly halfway between $\alpha$ and $\frac{\pi}{2}$ i.e. $\frac{\frac{\pi}{2} - \alpha} {2} = \frac{\pi}{4} - \frac{\alpha}{2}$
A: On a normal ground-to-ground projection, the angle for maximum range is π/4.
Intuitively, for an inclined plane, you would think that the angle for maximum range would be the angle θ that makes a π/4 angle with the ground on top of the α of the inclined plane.

That would be true if the projectile landed back on the ground and not on the incline.


But since the projectile does not fall back on the ground, I think of θ as the angle that makes π/4 with the angle bisector of α (the angle the incline plane makes with the ground)



