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JEE Mains · Physics · STD 11 - Motion in straight line
  • JEE Mains 2022
  • Hard
  • MCQ

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.

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Answer & Solution

Correct Answer

(B) \(\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}\)

Step-by-step Solution

Detailed explanation

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 \(t _{1} \& t _{2}\) )

\(\frac{t_{2}}{t_{1}}=\frac{\sqrt{2 g h}+\sqrt{2 g h-4 \times \frac{g}{2} \times \frac{h}{3}}}{\sqrt{2 g h}-\sqrt{2 g h-4 \times \frac{g}{2} \times \frac{h}{3}}}=\frac{\sqrt{2 g h}+\sqrt{\frac{4 g h}{3}}}{\sqrt{2 g h}-\sqrt{\frac{4 g h}{3}}}=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\)

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