InterviewSolution
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InterviewSolution
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A particle is projected from ground with speed u and at an angle `theta` with horizontal. If at maximum height from ground, the speed of particle is `1//2` times of its initial velocity of projection, then find its maximum height attained.A. `(u^(2))/(g)`B. `(2u^(2))/(g)`C. `(u^(2))/(2g)`D. `(3u^(2))/(8g)` |
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Answer» Correct Answer - D Given at maximum height `u cos theta =(1)/(2) u rArr cos theta = (1)/(2) :. Theta = 60^(@)` `:. H = (u^(2) sin^(2) theta)/(2g) = (u^(2) sin^(2) 60^(@))/(2g) = (3u^(2))/(8g)` |
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