A ball is thrown up vertically with a certain velocity so that it reaches a maximum height h. Find the ratio of the times in which it is at height $$\frac{h}{3}$$ while going up and coming down respectively.

We have a ball thrown vertically upward reaching maximum height $$h$$. At the top, all kinetic energy is converted to potential energy, so $$\frac{1}{2}mu^2 = mgh$$, giving $$u = \sqrt{2gh}$$.

Using the equation $$s = ut - \frac{1}{2}gt^2$$ with $$s = \frac{h}{3}$$, we get $$\frac{h}{3} = ut - \frac{1}{2}gt^2$$. Since $$u = \sqrt{2gh}$$ and $$h = \frac{u^2}{2g}$$, we substitute $$\frac{h}{3} = \frac{u^2}{6g}$$. The equation becomes $$\frac{u^2}{6g} = ut - \frac{1}{2}gt^2$$, which rearranges to $$3g^2t^2 - 6gtu + u^2 = 0$$. Dividing through by $$g^2$$ and writing $$T_0 = \frac{u}{g}$$ (total time to reach the top), we get $$3t^2 - 6T_0 t + T_0^2 = 0$$.

Applying the quadratic formula: $$t = \frac{6T_0 \pm \sqrt{36T_0^2 - 12T_0^2}}{6} = \frac{6T_0 \pm \sqrt{24T_0^2}}{6} = \frac{6T_0 \pm 2\sqrt{6}\,T_0}{6} = T_0\left(\frac{3 \pm \sqrt{6}}{3}\right)$$.

The ball is at height $$\frac{h}{3}$$ at two times: $$t_1 = T_0\left(\frac{3 - \sqrt{6}}{3}\right)$$ (going up) and $$t_2 = T_0\left(\frac{3 + \sqrt{6}}{3}\right)$$ (coming down).

The required ratio is $$\frac{t_1}{t_2} = \frac{3 - \sqrt{6}}{3 + \sqrt{6}}$$. Rationalising by multiplying numerator and denominator by $$(3 - \sqrt{6})$$: $$\frac{(3 - \sqrt{6})^2}{9 - 6} = \frac{9 - 6\sqrt{6} + 6}{3} = \frac{15 - 6\sqrt{6}}{3} = 5 - 2\sqrt{6}$$.

Now we check Option B: $$\frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}}$$. Rationalising: $$\frac{(\sqrt{3} - \sqrt{2})^2}{3 - 2} = (3 - 2\sqrt{6} + 2) = 5 - 2\sqrt{6}$$.

This matches our result exactly. Hence, the correct answer is Option B.