Question

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.

• A. $\frac{\sqrt{2}-1}{\sqrt{2}+1}$
• B. $\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}$
• C. $\frac{\sqrt{3}-1}{\sqrt{3}+1}$
• D. $\frac{1}{3}$

Answer 1

Answer: (B)

$ \text { Max. Height }= h =\frac{ u ^2}{2 g } $
$ \Rightarrow u =\sqrt{2 gh }$
$ S = ut +\frac{1}{2} at ^2$
$ \frac{ h }{3}=\sqrt{2 ght }+\frac{1}{2}(- g ) t ^2$
$ \frac{ gt ^2}{2}-\sqrt{2 ght }+\frac{ h }{3}=0$
(Roots are $t _1 \& t _2$ )
$\frac{ t _2}{ t _1}=\frac{\sqrt{2 gh }+\sqrt{2 gh -4 \times \frac{ g }{2} \times \frac{ h }{3}}}{\sqrt{2 gh }-\sqrt{2 gh -4 \times \frac{ g }{2} \times \frac{ h }{3}}}$
$=\frac{\sqrt{2 gh }+\sqrt{\frac{4 gh }{3}}}{\sqrt{2 gh }-\sqrt{\frac{4 gh }{3}}}=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}$