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# A ball of mass 50 g is dropped from a height of 20 m.A boy on the ground hits the ball vertically upwards with a bat with an average force of 200 N, so that it attains a vertical height of 45 m. The time for which the ball remains in contact with the bat is [Take g= 10 m/s2]

A
1/20th of a second
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B
1/40th of a second
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C
1/80th of a second
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D
1/120th of a second
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Solution

## The correct option is C 1/80th of a secondVelocityoftheballjustbeforehittingthebat=V1V12=02+2gs=2×10×20=400V1=−20m/s,V1isdownwardsheightattainedbytheballafterbeinghit:−45mThespeedjustafterbeinghitbythebat=V2⇒0=V22−2gs⇒V22=2×10×45=900V2=30m/sec(upwards)Changeinvelocity=V2−V1=30−(−20)=50m/sChangeinmomentum=mV2−mV1=mΔV=501000×50=2.5kgm/sforce=ΔpΔt=200NΔt=2.5200=180secHence,option (C) is correct answer.

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