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A particle is projected at point A from ...

A particle is projected at point A from an inclination plane with inclination angle `theta` as shown in figure. The magnitude of projection velocity is `vecu` and its direction is perpendicular to the plane. After some time it passes from point B which is in the same horizontal level of A, with velocity `vecv`. Then the angle between `vecu and vecv` will be

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The correct Answer is:
`[u = [(gR(1+3sin^(2)beta))/(2sin beta)]^(1//2)]`
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