Projectile Motion — Horizontal Range on Level Ground For a projectile launched with initial speed u at an elevation angle θ above the horizontal (neglecting air resistance), what is the horizontal range R on level ground?
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AR = (u^2 * sin 2θ) / g
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BR = (2 u^2 * sin θ * cos θ) / g^2
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CR = (u * sin 2θ) / g
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DR = (u^2 * cos 2θ) / g
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Answer
Correct Answer: R = (u^2 * sin 2θ) / g
Explanation
Introduction / Context: Horizontal range is a fundamental result in projectile motion used in ballistics, sports, and engineering. On level ground without drag, the closed-form expression depends on the launch speed u, launch angle θ, and gravity g.
Given Data / Assumptions:
- Launch speed u and angle θ.
- Flat ground; same elevation launch and landing.
- No air resistance; constant gravitational acceleration g.
Concept / Approach: Resolve motion into horizontal and vertical components, determine time of flight from vertical motion, and multiply by horizontal speed to obtain the range. Use the double-angle identity to simplify.
Step-by-Step Solution:
Horizontal speed: u_x = u cos θ (constant). Time of flight: T = 2 u sin θ / g (from y = 0 symmetry). Range: R = u_x * T = (u cos θ) * (2 u sin θ / g) = (u^2 * sin 2θ) / g.Verification / Alternative check: Maximum range occurs at θ = 45°, giving R_max = u^2 / g, consistent with differentiating R(θ) for optimum.
Why Other Options Are Wrong: Option (b) has g^2 in the denominator; (c) incorrectly omits a factor u; (d) uses cos 2θ which is not derived from the horizontal-range calculation.
Common Pitfalls: Forgetting the same-level assumption; confusing sin 2θ with 2 sin θ or not applying the double-angle identity correctly.
Final Answer: R = (u^2 * sin 2θ) / g.