Maximum range condition (ideal projectile without air resistance): For a given launch speed on level ground, which projection angle gives the maximum horizontal range?

Introduction / Context:In ideal projectile motion (no air drag, level launch and landing), the range depends on the sine of twice the launch angle. Identifying the angle that maximizes this trigonometric factor is a classic optimization in elementary mechanics.

Given Data / Assumptions:

• Constant launch speed u; flat terrain with same launch and landing elevations.
• No air resistance; constant gravitational acceleration g.

Concept / Approach:The range on level ground is R = (u^2 / g) * sin(2θ). For fixed u and g, maximizing R is equivalent to maximizing sin(2θ). The sine function attains its maximum value of 1 at 2θ = 90°, i.e., θ = 45°.

Step-by-Step Solution:

Verification / Alternative check:Angles θ and (90° − θ) yield the same range (since sin(2θ) = sin(180° − 2θ)); among practical angles, 45° is the symmetric mid-point providing maximum range.

Why Other Options Are Wrong:

• 30° and 60° produce the same range but less than the maximum.
• “None” is incorrect under the ideal assumptions.

Common Pitfalls:Forgetting the effect of air resistance (which can shift the optimal angle below 45° in real life), or confusing elevation differences which modify the formula.

Final Answer:45°