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

V = u cos θ sec β

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V = u cos β sec θ

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V = u cos θ

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V = u sec β

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Solution

The correct option is A $V = u cos θ sec β$
No acceleration is there in the horizontal direction so the horizontal component will always be the same.

$V_{h} = u C o s θ$

If angle of net velocity is $β$ with horizontal so $T a n β = \frac{V_{v}}{V_{h}}$ , $V_{v} = V sin β$ and $V_{h} = u C o s θ$

So $T a n β = \frac{V S i n β}{u C o s θ} => V = u C o s θ . s e c β$

So the best option is option A.

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