JEE Mains · Physics · STD 11 - 2. motion in straight line
A tennis ball is dropped on to the floor from a height of \(9.8\,m\). It rebounds to a height \(5.0\,m\). Ball comes in contact with the floor for \(0.2\,s\). The average acceleration during contact is \(..........ms ^{-2}\). [Given \(g =10\,ms ^{-2}\) ]
- A \(120\)
- B \(121\)
- C \(122\)
- D \(189\)
Answer & Solution
Correct Answer
(A) \(120\)
Step-by-step Solution
Detailed explanation
\(v _{ i }=\sqrt{2 gh _{ i }}\) \(=\sqrt{2 \times 10 \times 9.8} \downarrow\) \(=14 m / s \downarrow\) \(v _{ f }=\sqrt{2 gh _{ f }}\) \(=\sqrt{2 \times 10 \times 5} \uparrow\) \(= 1 0 ~ m / s \uparrow\)…
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