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A particle is projected with a velocity u. So that its horizontal range and maximum height reached are equal. The maximum height reached is

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${2u}^{2} / 3g$

${4u}^{2} / 5g$

$u^{2} / g$

${8u}^{2} / 17g$

answer is D.

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Detailed Solution

$R = H_{\max}$

$2 u^{2} \frac{\sin θ \cos θ}{g} = \frac{u^{2} \sin^{2} θ}{2 g}$

$\tan θ = 4$

$\sin θ = \frac{4}{\sqrt{17}}$

$H_{\max} = \frac{u^{2} .16 / 17}{2 g} = \frac{8 u^{2}}{17 g}$

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