I assume that the ball is stationary (v=0) at point B, so its total energy is just potential energy, and it is equal to 7.35 J.
At point A, all this energy has converted into kinetic energy, which is:
[tex]K= \frac{1}{2}mv^2 [/tex]
And since K=7.35 J, we can find the velocity, v:
[tex]v= \sqrt{ \frac{2K}{m} }= \sqrt{ \frac{2 \cdot 7.35 J}{1.5 kg} }=3.1 m/s [/tex]