How much energy is dissipated in the first bounce?

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1. Oct 25, 2016

emily081715

1. The problem statement, all variables and given/known data
A 0.70-kg basketball dropped on a hardwood floor rises back up to 66 % of its original height. If the basketball is dropped from a height of 1.4 m , how much energy is dissipated in the first bounce?

2. Relevant equations
U=mgy

3. The attempt at a solution
U=(0.70)(9.8)(1.4)
=9.604 J
i'm not sure where the error is

2. Oct 25, 2016

olivermsun

Now, can you tell how much energy the basketball had after the first bounce?

3. Oct 25, 2016

emily081715

how do i find how much energy it starts with, i thought that would be the potential energy? would i need to find energy by adding potential and kinetic, cause if so, i don't have a speed

4. Oct 25, 2016

olivermsun

Typically in these problems if they say the ball was "dropped" they mean it was just released with 0 initial speed.

So what about the info you are given regarding the energy after the first bounce?

5. Oct 25, 2016

emily081715

wouldn't i still need the final energy to calculate the kinetic energy of the first bounce? after the bounce is 66% of the original height so 0.924m

6. Oct 25, 2016

olivermsun

You have the max height after the first bounce, so that tells you the potential energy. What is the kinetic energy at the very moment that the max height is achieved?

7. Oct 25, 2016

emily081715

0?

8. Oct 25, 2016

olivermsun

Sure, so you have all the information you need.

9. Oct 25, 2016

emily081715

how?

10. Oct 25, 2016

emily081715

do i just subtract potential energies from the start and the max of second bounce

11. Oct 25, 2016

olivermsun

This part I leave to you to consider.

12. Oct 25, 2016

emily081715

the answer is 3.3 thanks for walking me through