InterviewSolution
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InterviewSolution
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The speed of a projectile at its maximum height is `sqrt3//2` times its initial speed. If the range of the projectile is n times the maximum height attained by it, n is equal to :A. `(4)/(3)`B. `2sqrt(3)`C. `4sqrt(3)`D. `(3)/(4)` |
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Answer» Correct Answer - C Let u be the initial speed and `theta` is the angle of the projection. Speed at the maximum height, `v_(H)=ucostheta=(sqrt(3))/(2)u` `:.costheta=(sqrt(3))/(2)ortheta=cos^(-1)((sqrt(3))/(2))=30^(@)` Range, `R=(u^(2)sin2theta)/(g)` and maximum height, `H=(u^(2)sin^(2)theta)/(2g)` As R=PH(Given) `:.(u^(2)sin2theta)/(g)=P(u^(2)sin^(2)theta)/(2g)` `2sinthetacostheta=(P)/(2)sin^(2)thetaortantheta=(4)/(P):.P=(4)/(tan30^(@))=4sqrt3` |
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