What is the time of flight T of a projectile launched with initial velocity u at angle alpha?
Answer and explanation
Correct answer: B. 2u sin alpha / g
- Step 1: Vertical displacement is y = (u sin alpha)t - (1/2)g t^2. Setting y = 0 at landing. Step 2. t [u sin alpha - (1/2)g t] = 0 => T = 2u sin alpha / g. Answer: 2u sin alpha / g. Why not the others: u sin alpha / g is time to peak; u^2 sin alpha / g has wrong dimensions; 2u cos alpha / g uses horizontal speed.
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