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A body is projected with velocity $v_{1}$ from the point A as shown in figure
At the same time, another body is projected vertically upwards from B with velocity $v_{2} .$ The point B lies vertically below the highest
point. For both the bodies to collide, $\frac{v_{2}}{v_{1}}$ should be

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0.5

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

answer is B.

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Detailed Solution

Two bodies will collide at highest point if both cover same vertical height in the same time. So,

$\frac{V_{1}^{2} \sin^{2} 30}{2 g} = \frac{V_{2}^{2}}{2 g} \Rightarrow$

$\Rightarrow \frac{V_{2}}{V_{1}} = \sin 30 = 0.5$

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