@ramyani: The problem asks for the minimum velocity so that the block does not tip over. In your solution you were attempting to see that the block does not even leave the ground in the first place.
If the velocity is sufficiently high, the block begins to rotate about hinge AB with gravity providing the retarding moment. The extremal condition is when the centre of mass is directly above the hinge. Since the torque is zero, the angular acceleration too is zero at this point. So, if you have angular velocity also become zero, the block cannot tip over.
So now apply the energy equation. The non-conservative forces make no contribution to the equation. At the final point, the KE is zero. The change in potential energy of the block can be calclulated by seeing how much the centre of mass shifts in the vertical direction from y = a to y =

2a (which is where I made a monumental error!).
Only as I noted before, I am not convinced that you can apply conservation of angular momentum at the beginning with gravity watching us. Could any of you please clear this point with your teachers as the text book seems to agree with this interpretation?
Some other clarifications to points raised in this thread: You definitely cannot use momentum conservation, the reaction at the hinge prohibits this. Normal Reaction (of the table ) bid us good bye the second the block took off. At the hinge these forces do no work.