$a)\;4 \\b)\;2 \\c)\;1 \\ d)\;0.5 $

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For Ball $B_2 : 0-v^2=-2gh$

$h=\large\frac{v^2}{2g}$

For Ball $B_1$ to reach the same height the vertical component of velocity is $u \sin \theta=u \sin 30$

$\quad \qquad= u/2$

Therefore $\large\frac{-u^2}{4}$$=-2g\bigg(\large\frac{v^2}{2g}\bigg)$

$\qquad=\large\frac{u^2}{v^2}$

$\qquad=4$

$\quad\large\frac{u}{v}$$=2$

Hence b is the correct answer.

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