Projectile launched at an angle to the vertical: angle for maximum horizontal range A projectile is fired at angle θ to the vertical (so its angle to the horizontal is 90° − θ). Neglecting air resistance and assuming level ground, for which value of θ will the horizontal range be maximum?

Introduction / Context:
The range of a projectile on level ground depends on the sine of twice the launch angle measured from the horizontal. When the angle is given relative to the vertical, a simple trigonometric adjustment is needed to find the maximizing condition.

Given Data / Assumptions:

• No air resistance; same launch and landing elevation.
• Initial speed fixed.
• Angle to the vertical is θ, so the angle to the horizontal is α = 90° − θ.

Concept / Approach:

Range R on level ground is R ∝ sin(2α), where α is measured from the horizontal. Since α = 90° − θ, we can express R in terms of θ and maximize it by using properties of the sine function.

Step-by-Step Solution:

Verification / Alternative check:

Symmetry of complementary angles confirms maximum range occurs for a 45° elevation from the horizontal, which corresponds to 45° from the vertical as well.

Why Other Options Are Wrong:

0° and 90° give zero range; 30° or 60° yield sin(60°) = sin(120°) = √3/2, which is less than 1.

Common Pitfalls:

Forgetting to convert between angle-to-vertical and angle-to-horizontal; maximizing sin θ instead of sin 2θ.

Final Answer:

45°