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Question

# A particle is thrown with a speed u at an angle θ with the horizontal. When the particle makes an angle β with the horizontal its speed changes to V. Then

A
V=ucosθsecβ
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B
V=ucosβsecθ
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C
V=ucosθ
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D
V=usecβ
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Solution

## The correct option is A V=ucosθsecβNo acceleration is there in the horizontal direction so the horizontal component will always be the same.Vh=uCosθIf angle of net velocity is β with horizontal so Tanβ=VvVh, Vv=Vsinβ and Vh=uCosθ So Tanβ=VSinβuCosθ=>V=uCosθ.secβSo the best option is option A.

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