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InterviewSolution
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35. A ball is dropped on the floor from a height of( 50 m10 m. It rebounds to a height of 2.5 m. If the ballis in contact with floor for 0.01 s, then theaverage acceleration during contact is nearly |
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Answer» For V before hitting the ground, v^2=u^2+2gh=0^2+2×9 8×10(it us preferable here to use g=9.8) Therefore v=14ms Now after bounce, V^2=U^2+2gH(here g is -ve) Hence V=-7m's Now a=(v-V)/t a=2100ms When it is dropped from 10m,Initial height = 10minitial velocity = 0velocity just before hotting ground =√2gh = √2*9.8*10 = 14.07 m/s (downward) after rebound,maximum height reached = 2.5mfinal velocity at top = 0initial velocity(just after rebound) =√2gh =√2*9.8*2.5 =√49 = 7 m/s (upward) assuming downward as positive directionSo velocity just before hitting ground = +14.07 m/svelocity just after hitting ground = -7 m/schange in velocity = +14.07 - (-7) = 21.07 m/stime = 0.01sacceleration = change in velocity/time = 21.07/0.01 = 2107 m/s² ~ 2100m/s² |
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