A 2kg ball falls from a height of H=12. What will be the balls maximum height af

Question

A 2kg ball falls from a height of H=12. What will be the balls maximum height after it bounces twice if it loses 1/3 of its velocity after each bounce?

Part 2: if the ball in the previous question is in contact with the floor for 0.10 seconds during the first bounce and 0.25 seconds during the second bounce, what is the average forced exerted by the floor on the ball during each bounce?

Explanation / Answer

1. find the velocity from energy conservation mgh=mv^2/2 --> v0=sqrt(2gh)=15.33 m/sec after the first bounce v1=v0*2/3=10.22 msec after the second bounce v2=v1*2/3=6.81 msec find the height from energy conservation mgh=mv^2 --> h=v2^2/2g=2.37 m 2. F=dP/dt p0=mv0=2*15.33=30.66 --> F=30.66/0.1=306.6 N p1=mv1=2*10.22=20.44 --> F=20.44/0.25=81.76 N

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