# A different bouncy ball question...

1. Nov 10, 2015

### SHO-NUFF

So I've seen all over the internet about the efficiency of a bouncy ball and how high will it bounce on the 2nd, 3rd, 4th, etc. bounces? That's easy to figure out. Now I was thinking about it for a while and the question that hit me was how much energy do I need to add to the initial drop in order for the ball to return to the exact original height? I have a table that's 1.03 m high, the ball has a mass of 7.362 g and a density of 0.90 g/cm^3. The ball is bouncing off a concrete floor that, from what I can find, has a density of 2,400 kg/cm^3. The ball also take approximately 0.5 seconds to fall and impact the floor and approximately another 0.25 seconds to reach the height of the first bounce of about 0.835 m.

I'm not sure if all this information is all relevant but it's what I've observed by simply bouncing the ball a couple dozen times. Please someone help point me in the right direction here. Again, what I'm looking for is the amount of energy required to be added in order for the ball to return to the exact original drop height on the first bounce. I'm missing something, overthinking it, or both.

2. Nov 10, 2015

### Khashishi

Using an Earth constant gravity approximation, the height is proportional to the potential energy of the ball. If the ball height is 1.03m before the first bounce and 0.835m after the first bounce, the fraction of energy retained after a bounce is
0.835/1.03 = 0.811
So, to bounce up to a height of 1.03m, you need to drop it from a higher height
1.03m / 0.811 = 1.27m
The potential energy is mgh, so the extra energy to lift to 1.27m is
W = mg(1.27m-1.03m)
(You shouldn't round the quantities until the end.)

3. Nov 10, 2015

### SHO-NUFF

Khashishi,

Thanks for the tip on rounding. What if I wanted to bounce it from a given height (in this case 1.03 m) and have it return to the exact height it was released from though?

4. Nov 10, 2015

### Khashishi

Calculate the difference in potential energy as I have shown above, and apply that energy to the initial kinetic energy of the ball.

5. Nov 10, 2015

### SHO-NUFF

Thanks for the help!!! My apologies for not getting the first time around. I'm new to physics.