Question 161 of 308

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 $\dfrac{h}{3}$ while going up and coming down respectively.

A$\dfrac{\sqrt{2}-1}{\sqrt{2}+1}$B$\dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}$C$\dfrac{\sqrt{3}-1}{\sqrt{3}+1}$D$\dfrac{1}{3}$

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Answer: (B) $\dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}$

The maximum height of a projectile thrown vertically upward is given by, $h=\dfrac{{u}^{2}}{2g}$.
$\Rightarrow u=\sqrt{2gh}$
Using equation of motion with constant acceleration,
$s=ut+\dfrac{1}{2}a{t}^{2}$
$\Rightarrow \dfrac{h}{3}=\sqrt{2gh}t-\dfrac{1}{2}g{t}^{2}$
$\Rightarrow \dfrac{g{t}^{2}}{2}-\sqrt{2ght}+\dfrac{h}{3}=0$
Solving the above quadratic equation we will get the times ${t}_{1}$ and ${t}_{2}$. $\Rightarrow \dfrac{{t}_{2}}{{t}_{1}}=\dfrac{\sqrt{2gh}+\sqrt{2gh-4\times \dfrac{g}{2}\times \dfrac{h}{3}}}{\sqrt{2gh}-\sqrt{2gh-4\times \dfrac{g}{2}\times \dfrac{h}{3}}}=\dfrac{\sqrt{2gh}+\sqrt{\dfrac{4gh}{3}}}{\sqrt{2gh}-\sqrt{\dfrac{4gh}{3}}}=\dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}$
Thus, the ratio $\dfrac{{t}_{1}}{{t}_{2}}=\dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}$