A

When analyzing projectile motion on an inclined plane, we want to find the angle of projection () that maximizes the range along the incline. Using the formula for range on an inclined plane, we can derive the condition for maximum range.
Step 1: The range formula is given by:
$$ R = \frac{v^2 \sin(2\alpha)}{g (\cos \theta + \sin \theta)} $$
where $R$ is the range, $v$ is the initial velocity, $g$ is the acceleration due to gravity, $ heta$ is the angle of the incline, and $rac{\sin(2\alpha)}{\cos \theta + \sin \theta}$ represents the effect of projection angle and incline on the range.
Step 2: To find the angle of projection for which the range is maximized, differentiate $R$ with respect to $ heta$ and set the derivative to zero.
Step 3: Solving this gives us the conditions for $ an(\alpha) = \frac{\sin(\theta)}{\cos(\theta)}$. Simplifying further leads us to conclude that:
$$ \alpha = \frac{\theta}{2} $$
This indicates that the range on the inclined plane is maximized when the angle of projection is half of the angle of inclination.
Therefore, the correct answer is option A.