A ball is thrown vertically upward with a certain velocity, reaching a maximum height $ h $. We need to find the ratio of the times it is at height $ \frac{h}{3} $ while ascending and descending, respectively.

The initial velocity of the ball $ v $ can be given by:

$ v = \sqrt{2gh} $

When the ball is at height $ \frac{h}{3} $, the equation of motion is:

$ \frac{h}{3} = \sqrt{2gh} \cdot t - \frac{1}{2}gt^2 $

Rearranging the equation, we get a quadratic equation in terms of $ t $:

$ \frac{g}{2} t^2 - \sqrt{2gh} \cdot t + \frac{h}{3} = 0 $

The ratio of the times taken to ascend and descend to the height $ \frac{h}{3} $ can be found using the quadratic formula solution for $ t $, considering the corresponding velocities:

$ \frac{t_1}{t_2} = \frac{\sqrt{2gh} + \sqrt{2gh - \frac{2gh}{3}} }{\sqrt{2gh} - \sqrt{2gh - \frac{2gh}{3}}} $

Simplifying the terms inside the fraction:

$ = \frac{\sqrt{2} + \frac{2}{\sqrt{3}}}{\sqrt{2} - \frac{2}{\sqrt{3}} } = \frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}} $