As per your syllabus, eqn 3.42b, the projection angle θ₀ that gives MAXIMUM range (for fixed v₀) is which value?
A90°
B60°
C30°
D45°
Answer & Solution
Correct answer: D. 45°
Your syllabus: sin 2θ₀ max = 1 at 2θ₀ = 90° → θ₀ = 45°.
Maximum range R_m = v₀²/g.
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