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(B) Let the horizontal distance covered by particle $A$ be $x_1$ and by particle $B$ be $x_2$. Particle $A$ moves with velocity $v_A = u/\sqrt{3}$ at

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$x/u$

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(B) Let the horizontal distance covered by particle $A$ be $x_1$ and by particle $B$ be $x_2$. Particle $A$ moves with velocity $v_A = u/\sqrt{3}$ at an angle of $30^\circ$ with the horizontal. Its horizontal component of velocity is $v_{Ax} = (u/\sqrt{3}) \cos 30^\circ = (u/\sqrt{3}) 	imes (\sqrt{3}/2) = u/2$. Particle $B$ moves with velocity $v_B = u$ at an angle of $60^\circ$ with the horizontal. Its horizontal component of velocity is $v_{Bx} = u \cos 60^\circ = u 	imes (1/2) = u/2$. Since both particles are moving towards each other,their relative horizontal velocity is $v_{rel} = v_{Ax} + v_{Bx} = u/2 + u/2 = u$. The initial horizontal separation between them is $x$. Time taken for the horizontal separation to become zero is $t = 	ext{Distance} / 	ext{Relative Velocity} = x / u$.

For the two given particles $A$ and $B$ as shown in the figure,how much time will it take for their horizontal separation to become zero?

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