classical-mechanicsnewtonian-mechanicsprojectile
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?
I think this is a good question to test if you think like a physicist. One thing physicist do is to analyze a problem by looking at limiting cases. Here there are two interesting cases: one where the angle of the ramp is zero, so that the ramp is actually just a flat level surface, the other is where the ramp becomes almost vertical (I would say the ramp actually being vertical doesn't make sense, but you can still take the limit).
Think about what the trajectory looks like when the angle is zero. Think what the trajectory becomes as you make the ramp more and more vertical. Then it should become clear what the times are in these limiting cases. Now if you trust the formula they gave you, you should have enough information from these two cases to determine the two quantities they want.
The law of reflection only holds if the collision is elastic and the object which the ball collides with does not move. If the collision is not elastic then the perpendicular component of velocity is smaller after than before, whereas the parallel component is the same. So when $e \lt 1$ the angle of reflection is greater than the angle of incidence.
The object being collided with must be very much more massive than the ball, so that it does not move. Momentum is conserved in the collision, provided the momentum of both objects is taken into account. If, for example, the ball is dropped onto and bounces from a wedge which is free to slide horizontally, the recoil of the wedge will increase the horizontal component of the velocity of the ball.
Friction and the spin of the ball will also affect the angle of reflection. During a finite collision time friction will reduce the parallel component of velocity. The spin of the ball affects whether there is relative motion between the ball and the plane during collision.
If the plane is fixed then the perpendicular and parallel components after the collision are $eV_0\cos i, V_0\sin i$ where $i=\theta$, so the angle of reflection is given by
$\tan r=\frac{V_0\sin i}{eV_0\cos i}=\frac{1}{e}\tan i$.
If the plane is a wedge which can slide horizontally then we can conserve momentum horizontally :
$mV_1\sin(i+r)=MU$.
The relative velocity of approach along the perpendicular is again $V_0\cos i$. The relative velocity of separation is $V_1\cos r+U\cos i$. Apply the Law of Restitution :
$V_1\cos r+U\cos i=eV_0\cos i$.
The relative velocity of approach along the parallel direction is initially $V_0\sin i$. If there is no friction between the ball and the wedge then this component is conserved in the collision, so
$V_0\sin i=V_1\sin r+U\sin i$.
The last 3 equations can be solved to find $V_1, r, U$.
Best Answer
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.