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JEE Main 2023 (29 Jan, Shift 1) · Motion Under Gravity

Motion Under Gravity question from JEE Main 2023 (29 Jan, Shift 1)

Q18JEE Main 2023 · 29 JAN · SHIFT 1Medium

Motion Under Gravity

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 0.2s. The average acceleration during contact is ___________ ms −2-2.

(Given g = 10 ms −2-2)

Numerical answer type. Work it out, then open the answer.

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Answer: 120

  1. Speed on hitting: v1=2gh1=2(10)(9.8)=14 m/sv_1 = \sqrt{2gh_1} = \sqrt{2(10)(9.8)} = 14\text{ m/s} (down)
  2. Speed on rebound: v2=2(10)(5)=10 m/sv_2 = \sqrt{2(10)(5)} = 10\text{ m/s} (up)
  3. a=ΔvΔt=10−(−14)0.2=240.2a = \frac{\Delta v}{\Delta t} = \frac{10 - (-14)}{0.2} = \frac{24}{0.2}
  4. a=120 m s−2a = 120\text{ m s}^{-2}
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