As shown in the figure, ABC = ACB = 60, so ABC is an equilateral triangle.

Therefore, the distance from the point light source to the ground AD=√3/2×BC=4√3(m).

Connect OB, OC. From the meaning of the question: ∠ Abe = 30 ⊙O is the inscribed circle of △ABE, so BO is the bisector of △ Abe, so △ ABO = 15.

∠OBC =∠ABC-∠ABO = 60- 15 = 45； Similarly, OCB = 45.

Then: △OBC is a right isosceles triangle.

Therefore, the distance from the center of the ball to the ground is OD=BC/2=4(m).