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A particle is projected from ground making an angle of $\tan^{- 1} \sqrt{2}$ with horizontal. Find the angle between its velocity vector and horizontal ground when its height above the ground will be equal to half of the maximum height attained by it.

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$60^{o}$

$30^{o}$

$45^{o}$

$37^{o}$

answer is C.

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Detailed Solution

Maximum height, $H = \frac{u^{2} \sin^{2} α}{2 g}$

where $α$ is the angle of projection.

At height $\frac{H}{2}, V \cos θ = u \cos α$

and ${(V \sin θ)}^{2} = {(u \sin α)}^{2} - 2. g . \frac{H}{2}$

$\Rightarrow {(V \sin θ)}^{2} = u^{2} \sin^{2} α - g (\frac{u^{2} \sin^{2} α}{2 g}) = \frac{u^{2} \sin^{2} α}{2}$

$\Rightarrow V \sin θ = u \sin α / \sqrt{2}$

$∴ \tan θ = \frac{V \sin θ}{V \cos θ} = \frac{\tan α}{\sqrt{2}} = \frac{\sqrt{2}}{\sqrt{2}} = 1 \Rightarrow α = 45^{o}$

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