JEE Mains · Physics · STD 11 - 5. work,energy,power and collision
A rubber ball is released from a height of \(5\, m\) above the floor. It bounces back repeatedly, always rising to \(\frac{81}{100}\) of the height through which it falls. Find the average speed of the ball. (Take \(g =10 ms ^{-2}\) ) (in \(ms ^{-1}\))
- A \(3.0\)
- B \(3.50\)
- C \(2.0\)
- D \(2.50\)
Answer & Solution
Correct Answer
(D) \(2.50\)
Step-by-step Solution
Detailed explanation
\(v _{0}=\sqrt{2 gh }\) \(v = e \sqrt{2 gh }=\sqrt{2 gh }\) \(\Rightarrow e =0.9\) \(S = h +2 e ^{2} h +2 e ^{4} h +\ldots \ldots \ldots\) \(t =\sqrt{\frac{2 h }{ g }}+2 e \sqrt{\frac{2 h }{ g }}+2 e ^{2} \sqrt{\frac{2 h }{ g }}+\ldots \ldots \ldots\)…
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