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  • A ball is thrown up with a certain velocity so that it reaches a height  Find the ratio of the two different times of the ball reaching <img alt="\frac{\mathrm{h}}{3}" src="/latex-image/?%5Cfrac%

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{\mathrm{h}}{3}  in both the directions.
Option: 1 \frac{\sqrt{2}-1}{\sqrt{2}+1}
Option: 2 \frac{1}{3}
Option: 3 \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}
Option: 4 \frac{\sqrt{3}-1}{\sqrt{3}+1}

Answers (1)

best_answer

V= \text { speed of projection}=\sqrt{2 g h}

Let at time t_{1}\; \text{and}\; t_{2} the ball reaches height h/3.

\begin{aligned} &s=u t+\frac{1}{2} a t^{2} \\ &\left(\frac{h}{3}\right)=\sqrt{2 g h} \times t+\frac{1}{2}(-g) t^{2} \\ &2 h=6 \sqrt{2 g h} t-3 g t^{2} \\ &3 g t^{2}-6 \sqrt{2 g h} t+2 h=0 \end{aligned}

\begin{aligned} t &=\frac{6 \sqrt{2 g h} \pm \sqrt{72 g h-4(3 g)(2 h)}}{2 \times 3 g} \\ &=\frac{6 \sqrt{2 g h} \pm \sqrt{48 g h}}{6 g} \end{aligned}

t=\frac{\sqrt{2 g h}(6 \pm \sqrt{24})}{6 g}

\begin{aligned} &t_{1}=\sqrt{\frac{2 h}{9}}\left(1-\sqrt{\frac{24}{36}}\right) \\ &t_{2}=\sqrt{\frac{2 h}{9}}\left(1+\sqrt{\frac{24}{36}}\right) \end{aligned}

\frac{t_{1}}{t_{2}}=\frac{ 1-\sqrt{\frac{24}{36}}}{1+\sqrt{\frac{24}{36}}}

\begin{aligned} &=\frac{1-\sqrt{\frac{2}{3}}}{1+\sqrt{\frac{2}{3}}} \\ \Rightarrow \frac{t_{1}}{t_{2}} &=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+2} \end{aligned}

The correct option is (3).

Posted by

vishal kumar

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