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The velocity of a projectile when it is at the greatest height is `(sqrt (2//5))` times its velocity when it is at half of its greatest height. Determine its angle of projection.A. `30^(@)`B. `45^(@)`C. `60^(@)`D. `37^(@)` |
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Answer» Correct Answer - C `(u cos alpha) = sqrt((2)/(5)) sqrt((u cos alpha^(2))+{(u sin alpha)^(2)-2gh))}` Here, `h = (H)/(2) = (u^(2)sin^(2)alpha)/(4g)` Solving this equation we get,`alpha = 60^(@)` |
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