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explain the answer 

Asked by Arushi Juyal 16th December 2017, 6:38 AM
Answered by Expert
Answer:
Vector OQ is drawn at O and parallel to Velocity vector v at P. If vector OQ makes an angle β with horizontal which is 90-α, then it can be seen from the figure that velocity vector v makes same angle β with the horizontal component. In projectile motion horizontal component of the velocity remains constant.
Hence v cos β = v cos (90-α) = v sin α = u cos α ;   hence v = u cot α
Answered by Expert 18th December 2017, 12:43 PM
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