A ball is dropped on to the floor from a height of 10 m. It rebounds to a height of 2.5 m. If the ball is in contact with floor for 0 .01 second, what is the average acceleration during contact ?
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a
700 m/s2
b
1400 m/s2
c
2100 m/s2
d
2800 m/s2
answer is C.
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Detailed Solution
According to above question,v1=(2gh)=2×10×10=200v2=−(2gh)=−2×10×2⋅5=−50So, Δv=v1−v2=(200)+(50)=350≈21Acceleration =ΔVΔt=210⋅01=2100m/s2.