Broken stand, falling ball

The left side of the picture is the view between the shorter tines, the right is a side view. The ball must be just the right size to have its center over a line joining the shorter tines. This is shown in the right-hand picture. The center of the ball is so high it is off the top of the picture, but the right triangles should be clear.

Using the left-hand picture to find d, the amount the ball dips below the short tines:

$d=r-\sqrt{r^{2}-5^{2}}$

Then using the right triangle in the right-hand picture to solve for $r$ :

$(5\sqrt{3})^{2}+(r-1-d)^{2}=r^2$

$75+(\sqrt{r^{2}-5^{2}}-1)^{2}=r^{2}$

$75+1-2\sqrt{r^{2}-25}+r^{2}-25=r^{2}$

$\sqrt{r^{2}-25}=\frac{51}{2}$

$r^{2}=\frac{2601}{4}+25=\frac{2701}{4}$

$r=\frac{\sqrt{2701}}{2}\approx\boxed{25.98557292}$