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?

Difficulty: Easy

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.

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