Projectile condition where range equals free-fall distance to reach the same launch speed For a projectile launched with initial speed u, let h be the vertical distance a particle must freely fall to acquire speed u (so h = u^2/(2g)). For what launch angle θ (with respect to the horizontal) will the horizontal range R be equal to h?

Introduction / Context:
This problem links kinematics of free fall to the range formula of projectile motion. It asks for the projection angle that makes the range equal to the vertical distance required to reach the same speed u by gravity alone.

Given Data / Assumptions:

• Initial speed u.
• Free-fall distance to reach speed u: h = u^2/(2g).
• Projectile launched and landing at the same elevation; no air resistance; constant g.

Concept / Approach:

On level ground, the horizontal range is R = (u^2/g) * sin(2θ). Set R = h and solve for θ. The equation reduces to sin(2θ) = 1/2, which has principal acute solution 2θ = 30°, giving θ = 15°.

Step-by-Step Solution:

Verification / Alternative check:

Numerical check: choose u = 10 m/s; h = 10^2/(2*9.8) ≈ 5.10 m. For θ = 15°, R = (100/9.8)sin 30° ≈ 10.200.5 ≈ 5.10 m, confirming equality.

Why Other Options Are Wrong:

30°, 45°, 60°, 75° correspond to sin(2θ) values of √3/2, 1, √3/2, and about 0.5 at different doubles; only θ = 15° makes sin(2θ) = 1/2 with 0° < θ < 45°.

Common Pitfalls:

Misreading the condition as maximum range; forgetting that the range contains sin(2θ), not sin θ; using degrees vs radians inconsistently.

Final Answer:

15°