JEE Main 2023PhysicsMotion In One DimensionEasyNumerical

JEE Main 2023 — Motion In One Dimension Question with Solution

JEE Main 2023 (29 Jan Shift 1)

Question

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 ______ m s-2. [Given g=10 m s-2]

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Show full solutionCorrect answer: 120
Correct answer
120

Step-by-step explanation

The speed of ball just before collision with ground is vi2=0+2ghi

 ⇒vi=2ghi

=2×10×9.8

=14 m s-1  Downward

The speed of ball just after collision is 0=vf2-2ghf

⇒vf=2ghf

=2×10×5

=10 m s-1  Upward

Average acceleration of ball is a→avg=Δv→Δt=vf--vit=10+140.2=240.2=120 m s-2.

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About this question

This is a previous-year question from JEE Main 2023, covering the Motion In One Dimension chapter of Physics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.