A particle is projected with a speed $V$ from a point O making an angle of $30^{\circ}$ with the vertical. At the same instant, a second particle is thrown vertically upwards with a speed $v$ from a point A. The two particles reach $H$ , the highest point of the parabolic path of particle one simultaneously. Then ratio $V / v$ is

3 \sqrt{2}

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2 \sqrt{3}

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\frac{2}{\sqrt{3}}

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\frac{\sqrt{3}}{2}

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Solution

The correct option is C $\frac{2}{\sqrt{3}}$
The maximum height reached by the particle 1 will be
$H = \frac{(V s i n 60^{\circ})^{2}}{2 g} . . . . . . . (1)$
The maximum height reached by the particle 2 will be
$H = \frac{v^{2}}{2 g} . . . . . . . (2)$
By (1) & (2)
$\frac{V^{2}}{v^{2}} {(\frac{\sqrt{3}}{2})}^{2} = 1$
$\frac{V}{v} = \frac{2}{\sqrt{3}}$

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