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Two projectiles A and B are launched with initial speeds in the ratio 1: √2 and reach the same maximum height. If projectile A is projected at 45° to the horizontal, what is the launch angle of projectile B?

1. 0°
2. 60°
3. 30°
4. 45°

Correct answer: 30°

Solution

Maximum height H = u^2 sin^2(theta)/(2g). Equal heights give uA^2 sin^2(45) = uB^2 sin^2(thetaB). With uB = sqrt(2) uA: uA^2 (1/2) = 2 uA^2 sin^2(thetaB) -> sin^2(thetaB) = 1/4 -> sin(thetaB) = 1/2 -> thetaB = 30 degrees.

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