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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.
$\frac{\sqrt{2}-1}{\sqrt{2}+1}$
$\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}$
$\frac{\sqrt{3}-1}{\sqrt{3}+1}$
$\frac{1}{3}$
Solution
Max. Height $= h =\frac{ u ^{2}}{2 g }$
$\Rightarrow u =\sqrt{2 gh }$
$S =u t+\frac{1}{2} a t^{2}$
$\frac{ h }{3}=\sqrt{2 gh } t +\frac{1}{2}(- g ) t ^{2}$
$\frac{ gt ^{2}}{2}-\sqrt{2 ght }+\frac{ h }{3}=0 \quad$ (Roots are $t _{1} \& t _{2}$ )
$\frac{t_{2}}{t_{1}}=\frac{\sqrt{2 g h}+\sqrt{2 g h-4 \times \frac{g}{2} \times \frac{h}{3}}}{\sqrt{2 g h}-\sqrt{2 g h-4 \times \frac{g}{2} \times \frac{h}{3}}}=\frac{\sqrt{2 g h}+\sqrt{\frac{4 g h}{3}}}{\sqrt{2 g h}-\sqrt{\frac{4 g h}{3}}}=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}$
