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The height of a particle projected with ...

The height of a particle projected with velocity u at an angle `alpha` with the horizontal is u `sinalpha.t-(1)/(2)"gt"^(2)` (g= constant) at any time t. Find the greatest height attained and the time of reaching it .

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The correct Answer is:
Greatest height `=(u^(2)sin^(2)alpha)/(2g)` and the time of reaching it `=(usinalpha)/(g)`
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