Let * * * have n small squares, and the mass of each square is m. When the lower end moves below A and the distance to A is 12L, the speed of the small squares reaches the maximum. At this time, the total resultant force is zero, which is obtained according to the equilibrium condition.

12? μnmgcosθ=nmgsinθ

Solution, μ=2tanθ.

Let the distance from the lower end of the small square to A be X, and when it stops, it is obtained according to the kinetic energy theorem.

nmg(2L+x)sinθ-μnmgcosθ？ L2-μnmgcosθ(x-L)=0

Substitute μ=2tanθ into the solution, and x = 3l.

So the answer is: 2tanθ, 3L.