Projectile Motion — Cartesian Trajectory Equation Which of the following correctly gives y as a function of x for a projectile launched with speed u at angle α above the horizontal (neglecting air resistance)?
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Ay = x tan α − (g x^2) / (2 u^2 cos^2 α)
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By = x tan α + (g x^2) / (2 u^2 cos^2 α)
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Cy = (u^2 sin 2α) / g − x
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Dy = (2 u^2 / g) sin α − x^2
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E—
Answer
Correct Answer: y = x tan α − (g x^2) / (2 u^2 cos^2 α)
Explanation
Introduction / Context: The Cartesian equation of a projectile’s trajectory is a parabola. Deriving y(x) helps solve range, height, and position problems without explicitly using time after elimination.
Given Data / Assumptions:
- Initial speed u, projection angle α above horizontal.
- Gravity g constant; no air resistance.
Concept / Approach: Resolve initial velocity: u_x = u cos α, u_y = u sin α. Use horizontal motion x = u cos α * t and vertical motion y = u sin α * t − (1/2) g t^2. Eliminate time t by t = x / (u cos α) to get the Cartesian form.
Step-by-Step Derivation:
x = u cos α * t ⇒ t = x / (u cos α). y = u sin α * t − (1/2) g t^2. Substitute t: y = x tan α − (g x^2) / (2 u^2 cos^2 α).Verification / Alternative check: At x = 0 ⇒ y = 0; the curvature term is negative (concave down), consistent with gravity. Units also match length on both sides.
Why Other Options Are Wrong: The “+” sign before the gravity term would make the path open upward; the other two options are unrelated forms and do not represent y(x) for projectile motion.
Common Pitfalls: Forgetting the cos^2 α in the denominator; sign errors on the gravity term; failing to eliminate time correctly.
Final Answer: y = x tan α − (g x^2) / (2 u^2 cos^2 α).