MHT CET · Physics · Laws of Motion
A ball is dropped on the floor from a height of 20 m . It rebounds to a height of 5 m . Ball remains in contact with floor for 1 s . The average acceleration during contact is
(acceleration due to gravity \(=10 \mathrm{~m} / \mathrm{s}^2\) )
- A \(30 \mathrm{~m} / \mathrm{s}^2\)
- B \(20 \mathrm{~m} / \mathrm{s}^2\)
- C \(40 \mathrm{~m} / \mathrm{s}^2\)
- D \(35 \mathrm{~m} / \mathrm{s}^2\)
Answer & Solution
Correct Answer
(A) \(30 \mathrm{~m} / \mathrm{s}^2\)
Step-by-step Solution
Detailed explanation
\(v_1 = -\sqrt{2gh_1} = -\sqrt{2 \times 10 \times 20} = -20 \text{ m/s}\) \(v_2 = \sqrt{2gh_2} = \sqrt{2 \times 10 \times 5} = 10 \text{ m/s}\)
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