Introduction:
In this question, we will discuss the motion of a uniform rod AB of mass m and length 2a when it is allowed to fall under gravity with AB in a horizontal position.

Analysis:
When the rod AB is allowed to fall under gravity with AB in a horizontal position, it will start rotating about its center of mass with an angular speed ω. The linear speed of the rod at any point along its length will be v = ωr, where r is the distance of the point from the center of mass.

The potential energy of the rod at the initial position is zero, and all of it gets converted into kinetic energy as the rod falls. Therefore,

mg(2a) = (1/2)mv^2 + (1/2)Iω^2,

where I is the moment of inertia of the rod about its center of mass.

The moment of inertia of the rod about its center of mass is given by

I = (1/12)ma^2 + (1/4)ma^2 = (1/3)ma^2.

Substituting the value of I in the above equation, we get

mg(2a) = (1/2)mv^2 + (1/2)(1/3)ma^2ω^2.

Simplifying the equation, we get

v = (2/3)√(ga).

Therefore, the speed of the rod is v = (2/3)√(ga).

Conclusion:
In conclusion, when a uniform rod AB of mass m and length 2a is allowed to fall under gravity with AB in a horizontal position, its speed is given by v = (2/3)√(ga).