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A particle of mass m is projected with a velocity `mu` at an angle of `theta` with horizontal.The angular momentum of the particle about the highest point of its trajectory is equal to :A. `("mu"^(3)sin^(2)thetacostheta)/(3g)`B. `(3"mu"^(3)sin^(2)thetacostheta)/(3g)`C. `("mu"^(3)sin^(2)thetacostheta)/(2g)`D. `(2"mu"^(3)sinthetacos^(2)theta)/(3g)` |
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Answer» Correct Answer - C |
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