As per your syllabus, eqn 3.41, the maximum height h_m of a projectile is which expression?
A(v₀ sin θ₀)² / (2g)
Bv₀² / (2g)
Cv₀² sin 2θ₀ / g
D(v₀ cos θ₀)² / (2g)
Answer & Solution
Correct answer: A. (v₀ sin θ₀)² / (2g)
syllabus eqn 3.41: h_m = (v₀ sin θ₀)² / (2g).
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