What is the maximum height H achieved by a projectile launched with speed u at angle alpha?
Answer and explanation
Correct answer: B. u^2 sin^2 alpha / (2g)
- Step 1: Using v_y^2 = u_y^2 - 2g H with vertical speed v_y = 0 at apex. Step 2. 0 = (u sin alpha)^2 - 2g H => H = u^2 sin^2 alpha / (2g). Answer: u^2 sin^2 alpha / (2g). Why not the others: u^2 sin 2alpha / g is the horizontal range; u sin alpha / g is time to apex; u^2 / (2g) is maximum height when fired vertically.
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