Projectile on an inclined plane — angle for maximum range: For a projectile launched up an inclined plane that makes an angle β with the horizontal, the range measured along the plane is maximum when the angle of projection α (measured from the horizontal) equals:

Introduction / Context:
Optimizing projectile range on an incline appears in mining, ballistics, and sports engineering. Unlike the horizontal ground case (optimum 45°), the incline tilts the landing surface, shifting the best launch angle.

Given Data / Assumptions:

• Inclined plane makes angle β with the horizontal.
• Launch speed is fixed; air resistance is neglected.
• Angle of projection α is measured from the horizontal.

Concept / Approach:
Resolve motion into components parallel and perpendicular to the plane or use standard x–y components and impose the landing condition y = x * tanβ (the plane line). Express range along the plane R_plane as a function of α and β, then maximize with respect to α for a fixed speed.

Step-by-Step Solution:
Equation of the plane: y = x * tanβ.Projectile: x(t) = u * cosα * t; y(t) = u * sinα * t − (1/2) * g * t^2.At impact: u * sinα * t − (1/2) * g * t^2 = u * cosα * t * tanβ.Solve for time t > 0, then get x and distance along plane: R_plane = x * secβ.Differentiating R_plane(α) with respect to α and setting derivative to zero yields optimum α = 45° + β/2.

Verification / Alternative check:
Check limiting cases: For β = 0° (horizontal ground), α = 45°, matching the well-known result. As β increases, the optimum α increases accordingly, which is physically sensible because the plane rises upward.

Why Other Options Are Wrong:

• β/2, 30° + β/2, 60° + β/2: Do not satisfy the derived extremum condition and give suboptimal ranges except at special, coincidental β values.

Common Pitfalls:

• Maximizing horizontal range instead of range along the plane; the objective function differs.
• Forgetting the secβ factor converting horizontal displacement to along-plane distance.

Final Answer:
45° + β/2