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.

<p>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.</p> <p>The initial velocity of the ball $ v $ can be given by:</p> <p>$ v = \sqrt{2gh} $</p> <p>When the ball is at height $ \frac{h}{3} $, the equation of motion is:</p> <p>$ \frac{h}{3} = \sqrt{2gh} \cdot t - \frac{1}{2}gt^2 $</p> <p>Rearranging the equation, we get a quadratic equation in terms of $ t $:</p> <p>$ \frac{g}{2} t^2 - \sqrt{2gh} \cdot t + \frac{h}{3} = 0 $</p> <p>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 $, considering the corresponding velocities:</p> <p>$ \frac{t_1}{t_2} = \frac{\sqrt{2gh} + \sqrt{2gh - \frac{2gh}{3}} }{\sqrt{2gh} - \sqrt{2gh - \frac{2gh}{3}}} $</p> <p>Simplifying the terms inside the fraction:</p> <p>$ = \frac{\sqrt{2} + \frac{2}{\sqrt{3}}}{\sqrt{2} - \frac{2}{\sqrt{3}} } = \frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}} $</p>