ExamBro
ExamBro
JEE Mains · Physics · STD 11 - Work,energy,power and collision
  • JEE Mains 2021
  • Hard
  • MCQ

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 \(ms ^{-1}\))

Pick an option to check your answer.
Verified Solution

Answer & Solution

Correct Answer

(D) \(2.50\)

Step-by-step Solution

Detailed explanation

\(v _{0}=\sqrt{2 gh }\)

\(v = e \sqrt{2 gh }=\sqrt{2 gh }\)

\(\Rightarrow e =0.9\)

\(S = h +2 e ^{2} h +2 e ^{4} h +\ldots \ldots \ldots\)

\(t =\sqrt{\frac{2 h }{ g }}+2 e \sqrt{\frac{2 h }{ g }}+2 e ^{2} \sqrt{\frac{2 h }{ g }}+\ldots \ldots \ldots\)

\(v _{ av }=\frac{ s }{ t }=2.5\, m / s\)

Same subject
Explore more questions on app