# Find Coefficient of Restitution with just height bounced

1. Feb 17, 2016

### x2017

1. The problem statement, all variables and given/known data
A ball is dropped from a height of 9.02 m. After hitting the ground, the ball then rebounds to a height of 1.78 m. What is the coefficient of restitution associated with the ball and ground impact?

2. Relevant equations
e=(V'-v')/(V-v)

3. The attempt at a solution
I thought maybe this was a trick question and you just do 9.02-1.78=7.24 and that was the answer but I am wrong... I need a bit of guidance on how to proceed with this question.

EDIT:
I just realized I should be using e=v'/v instead since there is only one ball... And I think I have realized that v' is -9.81? Not completely sure though because I also think that that might only be acceleration... CONFUSED lol
I thought that maybe I could use acceleration to find velocity, but I don't have the time....

2. Feb 17, 2016

### haruspex

Yes, that's the acceleration. What SUVAT equation do you know that relates acceleration, distance and speeds?

3. Feb 17, 2016

### x2017

Thanks for the reply, I always seem to forget about the SUVAT equations...

vf2-vi2=2(9.81)(9.02)

I used 9.02m and not 1.78m because it travelled down 9.02m when gravity was acting on it and I don't know what it's acceleration upwards is. Hopefully this is correct!

vf2-vi2=2(9.81)(9.02)
vf2-vi2=176.9724
vf-vi=13.30
Δv=13.30m/s

Is the above something I can do? So now I have the change in velocity of the ball...
But only on the way down correct? or is this all together?

4. Feb 17, 2016

### haruspex

You have found the landing speed, yes. A few doubtful steps along the way, though.
You should be careful with signs. If up is positive, the acceleration is -9.81m/s2, and the distance is -9.02m. That way you get a positive Δ(v2).
Secondly, you cannot go straight from vf2-vi2 to vf-vi by square rooting. You need to plug in the value for vi (0) first.

Now you need to find the rebound speed by the same method.

5. Feb 17, 2016

### x2017

How do I get the acceleration for the rebound speed though? Since it's travelling upwards it's not gravity right? And I don't have enough information to solve for acceleration with Δv/Δt.

6. Feb 17, 2016

### haruspex

why not? Does gravity stop acting on a body just because it is moving upwards?

7. Feb 17, 2016

### x2017

No, I guess gravity is causing it to slow down as it approaches its peak height! Okay I understand that part now, thanks!

I solved for the rebound speed and got 5.91m/s.

e=v'/v
e=5.91/13.30
e=0.44 which comes up as correct!

This turned out to be pretty simple, thanks for helping me understand how to go about it haruspex! :)

8. Feb 17, 2016

### haruspex

Well done.

By the way, if you work it symbolically instead of plugging in numbers you will find the relationship between e and the height ratio. You don't actually need to calculate the velocities.