Question

A particle is projected with a velocity , so that its range on a horizontal plane is twice the greatest height attained. If g is acceleration due to gravity, then its range is

Moderate

Solution

$H = \frac{u^{2} \sin^{2} θ}{2 g} and R = \frac{u^{2} \sin 2 θ}{g} Since, R = 2 H, so \frac{u^{2} \sin 2 θ}{g} = 2 \frac{u^{2} \sin^{2} θ}{2 g} or 2 \sin θcos θ = \sin^{2} θ or \tan θ = 2 \begin{aligned} ∴ R = u^{2} \frac{2}{g} \sin θcos θ \\ = \frac{2 u^{2}}{g} \frac{2}{\sqrt{5}} \frac{1}{\sqrt{5}} = \frac{4 u^{2}}{5 g} . \end{aligned}$

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