Question

A particle is projected with a velocity v, so that its horizontal range twice the greatest height attained. The horizontal range is

Moderate

Solution

Given, a particle having horizontal range is twice the greatest height attained by it, i.e. R = 2H
$\frac{v^{2} \sin 2 θ}{g} = \frac{2 v^{2} \sin^{2} θ}{2 g}$
$\Rightarrow 2 \sin θ \cos θ = \sin^{2} θ \Rightarrow \tan θ = 2$
$\Rightarrow \sin θ = \frac{2}{\sqrt{5}} \Rightarrow \cos θ = \frac{1}{\sqrt{5}}$
$∴ R = \frac{v^{2} \sin 2 θ}{g} = \frac{v^{2} \times 2 \sin θ \cos θ}{g}$
$\Rightarrow R = v^{2} \times \frac{2}{g} \times \frac{2}{\sqrt{5}} \times \frac{1}{\sqrt{5}}$
$i.e. R = \frac{4 v^{2}}{5 g}$

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