JEE Mains · Physics · STD 11 - 2. motion in straight line
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 & 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 \(t _{1} \& t _{2}\) )…
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