With two balls connected to a string find minimum upward velocity that can be given to one of the balls such that the other leaves the ground The exact question is given below:

Two identical small balls A and B each of mass m connected by a light inextensible cord of length l are placed on a frictionless horizontal floor. With what velocity u must the ball B be projected vertically upwards so that the ball A leaves the floor? Acceleration of free fall is g.
  
$$Answer: u \ge \sqrt{3gl} $$

I found this question in a book that is focused towards olympiads and was not able to figure it out. Now, even after many attempts have not been able to arrive at the answer in a completely satisfactory manner. I am looking for a satisfactory explanation and solution to this question.
 A: Let us consider the final state when $B$ is directly above $A$ which is still touching the ground. In order to lift up $A$, the tension in the string must be greater than the weight of $A$.
$$T\ge mg$$
But since ball $B$ will be in circular motion around $A$, the tension is equal to the centripetal force.
$$\frac {mv^2}l\ge mg$$
$$v\ge \sqrt {gl}$$
Finally, equating initial kinetic energy and final kinetic and potential energy. 
$$\frac 12mu^2=\frac 12 mv^2+mgl$$
$$u^2=v^2+2gl$$
$$u^2\ge gl+2gl$$
$$u\ge \sqrt {3gl}$$
A: As @saumy points out, both A and B see equal and opposite horizontal forces and therefore the horizontal velocities are equal and opposite at all times. When B is directly above A, both bodies are instantaneously rotating around the mid-point of the string.  The tension in the string due to the rotation is $mv^2/(L/2)$. When this force is greater than the weight of A, then A will lift off the table:
$$mv^2/(L/2) >= mg$$
Now equating the initial kinetic energy with the final kinetic + potential energy:
$$m_Bu^2/2 = m_Av^2/2 + m_Bv^2/2 + m_BgL$$
Finally setting $m=m_A=m_B$ and eliminating $v^2$, we get the limiting initial velocity:
$$u >= \sqrt{3gL}$$
That this is the same answer is probably not a coincidence.  However, it's not entirely clear to me what justifies saying this condition occurs when B is vertically above A. The rotation is faster and the tension in the string is higher just before it reaches this point. But I don't know whether the   vertical component of the tension is higher.
