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A ball of mass 0.120 kg is dropped from rest from a height of 1.25 m. It rebounds from the floor to reach a height of 0.600 m. What impulse was given to the ball by the floor in kg*m/s?

2006-10-23 14:39:45 · 1 answers · asked by activegirl 1 in Science & Mathematics Physics

1 answers

Just equate the kinetic and potential energies: .5mv^2=mgh
This means that v= sqrt(2gh)=sqrt(2*9.8*.6)=3.42m/s; therefore, the impulse is .12*3.43=.411.

2006-10-23 16:02:07 · answer #1 · answered by bruinfan 7 · 0 0

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