Question

A rubber ball is dropped on to flat ground from a height of 2.0m. It loses 10% of its kinetic energy at each bounce. Calculate how long it takes for the ball to first hit the ground and the time taken for the ball to come to rest.

Answer 1

so, you first need an equation to solve this.

Answer 2

and there's no time function.

Answer 3

I did the first part using h=1/2gt^2. The answer is 0.63s. I need help with the second part. O_o

Answer 4

Okay so figure out how much energy it had at 2.0 meters.
Then multiply that by \(1-0.10=0.90\) to get the amount or remaining energy. Find out how high it will go the next time.

Answer 5

and use the same method then to find out the time it will take to get that high... adding to it the time it took to hit the ground.

Answer 6

But wouldn't the no. of bounces reach infinity?

Answer 7

The ball comes to rest whenever it reaches it's highest point.

Answer 8

Because there is a sweet spot where velocity is 0.

Answer 9

Technically there is also a sweet spot right when it was let go and right when it bounced too...

Answer 10

Oh wait....
You want to know when the ball stops bouncing completely?
Sorry I did no correctly read into the question.

Answer 11

Is there a threshold of distance or something to where you consider it at rest?

Answer 12

Ohh, I think I get it... it may be that the ball bounces an infinite number of times and yet comes to rest because it is a convergent series.

Answer 13

Try to figure out the time it will take for the ball to go through \(n\) bounces, then do the limit to infinity.

Answer 14

@emcrazy14 Catch my drift?

Answer 15

We'll start with \(n=0\) when it hits the ground for the first time.
At this point is has \(E_0\) energy.
The formula for the energy after \(n\) bounces will be: \[
E(n) =E_0(0.90)^n
\]

Answer 16

Since \[
E = mgh
\]The height after \(n\) bounces will be: \[
h(n) = \frac{E(n)}{mg} = \frac{E_0(0.90)^n}{mg}
\]Now we need the time it will take to complete \(n\) bounces...

Answer 17

To get this we use \[
h=\frac{1/2}gt^2 \implies t = \sqrt{2hg}
\]But this is only the time it takes to do go up or down. To go both up or down we multiply by \(2\) \[
t = 2\sqrt{2hg}
\]This means after \(n\) bounces, the time it takes is: \[
t(n) = 2\sqrt{2gh(n)} = 2\sqrt{\frac{2E_0(0.90)^n}{m}}
\]Is this a convergent series... I wonder...
Maybe I'm not going about the problem right.

Answer 18

Oh wait, it's a geometric series... and \(|r|\leq 1\) of course it converges.

Answer 19

@emcrazy14 I'm not saying this is an easy problem, but it is doable.