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  • A ball is thrown up vertically with a certain velocity so that, it reaches a maximum height

A ball is thrown up vertically with a certain velocity so that, it reaches a maximum height \mathrm{h.}
Find the ratio of the times in which it is at height \mathrm{\frac{\mathrm{h}}{3}} while going up and coming down respectively.
 

Option: 1

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


Option: 2

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


Option: 3

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


Option: 4

\frac{1}{3}


Answers (1)

best_answer

\mathrm{S=ut+\frac{1}{2}at^{2}}

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

\mathrm{2 h=6 u t-3 g t^2}

\mathrm{3 g t^2-6 u t+2 h=0}

\mathrm{t=\frac{-(-6 u) \pm \sqrt{36 u^2-4 \times 3 g \times(2 h)}}{2 \times 39}}

\mathrm{t=\frac{6 u \pm \sqrt{36 u^2-24 g h}}{6 g} \rightarrow}(1)

\mathrm{v^2=u^2+2 a s}

\mathrm{0=u^2+2(-g)(+h)}

\mathrm{u^2=2 g h \rightarrow (2)}

From eq (1) and (2)

\mathrm{t=\frac{6 \sqrt{2 g h} \pm \sqrt{48 g h}}{6 g} }

\mathrm{t_1=\frac{6 \sqrt{2 g h}-\sqrt{48 g h}}{6 g} }

\mathrm{t_2=\frac{6 \sqrt{2 g h}+\sqrt{48 g h}}{6 g} }

\mathrm{\frac{t_1}{t_2}=\frac{6-\sqrt{24}}{6+\sqrt{24}}= \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}}

Hence 2 is correct option

Posted by

Kuldeep Maurya

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