As per your syllabus, eqn 3.40a, the time t_m to reach the maximum height of a projectile is which?
Av₀ / g
Bv₀ cos θ₀ / g
Cv₀ sin θ₀ / g
D2 v₀ sin θ₀ / g
Answer & Solution
Correct answer: C. v₀ sin θ₀ / g
syllabus eqn 3.40a: at max height v_y = 0 ⇒ t_m = v₀ sin θ₀ / g.
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