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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 enter image description here

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?

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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.

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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”.

enter image description here

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?

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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. $\frac{\frac{\pi}{2} - \alpha} {2} = \frac{\pi}{4} - \frac{\alpha}{2}$

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    $\begingroup$ -1 Not useful. This answer merely restates the result already given in the question. It does not not provide an intuitive reason why this rule has to be the case. $\endgroup$
    – sammy gerbil
    Commented Dec 3, 2017 at 20:40
  • $\begingroup$ @sammygerbil I believe the answer was useful to the OP, as can be seen by the fact that it's been accepted and commented. Also, I can't see why it is not an intuitive reason, as the result has to be properly demonstrated with the standard kinematics equations. $\endgroup$
    – Toool
    Commented Dec 4, 2017 at 17:27
  • $\begingroup$ Your answer does not explain why the angle is what it is. All you have shown is how the angle relates to the geometry. $\endgroup$
    – sammy gerbil
    Commented Dec 5, 2017 at 8:27
  • $\begingroup$ @sammygerbil At this point, I don't understand what you mean by an intuitive argument. Maybe you could show one ? Anyway, downvoting a correct answer that was accepted by the OP is quite useless. $\endgroup$
    – Toool
    Commented Dec 5, 2017 at 14:32
  • $\begingroup$ Your explanation is inductive not intuitive. Induction (in the loose sense) means : the rule works for case A $(\alpha=0)$ and it works for case B $(\alpha=\frac{\pi}{2})$ etc, therefore it probably works for all other cases in between. But it is possible that the rule only applies at these two extremes. An intuitive explanation shows why something must be true for all cases without giving a formal proof. eg The symmetry argument used in Why is the electric field of an infinite insulated plane of charge perpendicular to the plane? $\endgroup$
    – sammy gerbil
    Commented Dec 5, 2017 at 15:43
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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.

enter image description here

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)

enter image description here

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