# How long does it take the ball to come to a hault?

1. Jul 12, 2013

### iScience

the question is on here http://imgur.com/z4EFJiO

ignore the silly answer i found it on facebook and started to solve it just for fun and i ended up with...

http://imgur.com/pBkXZIs

sorry for the messy format, but i explained all my calculations (you're going to have to click to enlarge, and then click again to enlarge further.. again i apologize)

one embarrassing thing though, i don't know how to calculate the sum :( ... if anyone would be so kind.. would you demonstrate how?

2. Jul 12, 2013

### haruspex

How does the ball's horizontal speed change with each bounce?

3. Jul 13, 2013

### iScience

well, if the ball loses one joule each time, i assumed it was referring to the friction and the inelasticity of each bounce, together simplifying it for me saying that it loses 1J w/ each bounce so i assumed it was the 1J in the direction of the vectorial sum of v. ie... just the direction of v.

4. Jul 13, 2013

### haruspex

To me, the question is underspecified. There is no reason why it should lose energy in the same proportion in the two directions. In reality, I would expect most energy to be lost from the vertical component. Typically a ball stops bouncing long before it stops rolling. Forced to make an assumption, I would take it that all the lost energy comes from the vertical component. But that could be a wrong guess.

5. Jul 14, 2013

### iScience

assuming my assumption is correct, is my final equation right? i posted this thread because i make alot of mistakes and i wanted to check my answer

6. Jul 14, 2013

### haruspex

Inside the sum you have an x which I think should be i, right? The sqrt expression it is in then gives you the ball's speed immediately after i bounces. Then you subtract that from the vertical speed just before the first bounce? That can't be right.
Your assumption that the energy is lost in equal proportion from the vertical and horizontal components leads to a very nasty sum. I really don't think this is what is intended.

7. Jul 16, 2013

### iScience

yes, x(i)

not quite; the first sqrt, ie sqrt(196.2) is the ball's velocity(y) right after its first bounce. and then i subtract from this, the ball's subsequent losses in velocities(y).

just solving for fun... i honestly did have to assume sOMEthInG at the time due to the unspecific nature of the given problem, there were different assumptions i could've made, more likely ones that would be seen in nature, but i didn't see those at the time, er.. perhaps i did but anyhows i just went with this one :)

Last edited: Jul 16, 2013
8. Jul 16, 2013

### haruspex

That's not how I read it. It is certainly wrong. √196.2 is the vertical velocity you computed for just before the first bounce, and √596.2 is the total speed at that time. Try putting x = 0 or 1; you'll see that the contribution to total time goes negative.