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Question

To test the quality of a tennis ball, you drop it onto the floor from a height of 4.00m. It rebounds to a height of 2.00m. If the ball is in contact with the floor for 12.0ms, what is its average acceleration during that contact? Take g=10m/s2.

A
1.26×103ms2 (upward)
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B
1.26×103ms2 (downward)
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C
2.52×103ms2 (upward)
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D
2.52×103ms2 (downward)
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Solution

The correct option is A 1.26×103ms2 (upward)
Using, v2=u2+2as
Velocity of ball just before hitting the floor
v21=2×10×4
or v1=80m/s downwards
Velocity of ball after hitting the floor
0=v222×10×2
or v2=40m/s upwards

Using v=u+at for 12ms time (taking upwards direction as positive)
40=80+a×0.012
a=40+800.012
a=1272.4m/s21.26×103m/s2 upwards
Thus correct option is (a)

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