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A ball is thrown up with a certain velocity so that it reaches a height 'h'. Find the ratio of the two different times of the ball reaching $\frac{h}{3}$ in both the directions.

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$\frac{\sqrt{2} - 1}{\sqrt{2} + 1}$

$\frac{\sqrt{3} - 1}{\sqrt{3} + 1}$

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

$\frac{1}{3}$

answer is C.

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Detailed Solution

$u = \sqrt{2 g h}$

Now, $S = \frac{h}{3}, a = - g$

$S = u t + \frac{1}{2} a t^{2}$

$\frac{h}{3} = \sqrt{2 g h t} + \frac{1}{2} (- g) t^{2}$

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

From quadratic equation

$t_{1}, t_{2} = \frac{\sqrt{2 g h} \pm \sqrt{2 g h - \frac{4 g}{2} \frac{h}{3}}}{g}$

$\frac{t_{1}}{t_{2}} = \frac{\sqrt{2 g h} - \sqrt{\frac{4 g h}{3}}}{\sqrt{2 g h} + \sqrt{\frac{4 g h}{3}}} = \frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}}$

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