Projectile on an inclined plane – condition for maximum range In classical mechanics (neglecting air resistance), the range of a projectile thrown up along an inclined plane becomes maximum under a specific launch direction. The direction of projection should bisect the angle between the inclined plane and the vertical when the range is intended to be what?

Introduction / Context:
A standard result in projectile motion on an inclined plane is the condition that maximizes the distance traveled along the plane. This is useful in civil and defense applications where launch angles must be optimized for maximum reach on sloped ground.

Given Data / Assumptions:

• No air resistance; acceleration due to gravity is uniform and vertical.
• The plane is straight, making an angle θ with the horizontal.
• Launch speed is fixed; only launch direction is varied.

Concept / Approach:

For a projectile launched relative to an inclined plane, the component of initial velocity along the plane and the time of flight together determine the range on the plane. Using resolved axes parallel and normal to the plane yields the range expression. Optimizing this expression with respect to the launch direction gives the maximum-range condition.

Step-by-Step Solution:

Verification / Alternative check:

Using the standard transformation to axes parallel and perpendicular to the plane gives the same maximizing condition α = 45°, independent of the plane slope when the launch speed is fixed. Geometrically, the bisector of the angle between plane and vertical is indeed 45° to the plane.

Why Other Options Are Wrong:

(a) Zero range occurs only if the projectile does not travel along the plane, not under the bisector condition. (c) Minimum range is not obtained at the bisector. (d) The statement is definite; a correct option exists. (e) Range clearly depends on plane angle and launch angle.

Common Pitfalls:

Confusing the angle to the horizontal with the angle to the plane; forgetting that the plane–vertical angle is 90°, so the bisector implies 45° to the plane.

Final Answer:

to be maximum