# Homework Help: Bouncing Ball Quadratic Graph

1. Nov 4, 2007

### NKKM

I have a bouncing ball quadratic equation and know that B represents the initial velocity. My question is why does B the initial velocity increase with each bounce of the ball. It seems paradoxical to me, particularly if total energy is decreasing.

y=ax^2 + bx + C

I have attached a copy of the graph to this forum. Any help would be greatly appreciated. Thank you

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2. Nov 4, 2007

### learningphysics

Might be better if you upload the picture to an image sharing site, and link... approval takes a while. Try this site:

http://www.imagevimage.com/

3. Nov 4, 2007

### NKKM

thanks for the tip
http://www.imagevimage.com/images/2_Bouncing_Ball_Graph.jpg

I am still confused as to why the initial velocity for each bounce appears to increase.

In addition, in the quadratic equation: c should be equal to initial displacement? But the numbers are so large.. I'm really not sure what is going on here

Thanks again

Last edited: Nov 4, 2007
4. Nov 4, 2007

### learningphysics

are you calculating the velocity at the right points?

you won't get the velocity at the point where it bounces by looking at the b-value...

you need to take the derivative of the equation... and plug in the time.

5. Nov 4, 2007

### NKKM

well what i need to do first is figure out what the coefficients in the equation of the graph mean:

so a = 1/2 * gravitational acceleration

b = initial velocity of the ball - why does this increase with each bounce; the question states that it should increase but I don't understand why?

c = ? - maximum height? but that does not make sense

6. Nov 4, 2007

### learningphysics

You need to plug in the right time... for example in the first fit equation... plug in x = 0.8... there you will get y = 0.

If the bounce was occuring at t = 0... then you could just look at the c-value... but the graph is shifted to the right.

7. Nov 4, 2007

### NKKM

well what i need to do first is figure out what the coefficients in the equation of the graph mean:

so a = 1/2 * gravitational acceleration

b = initial velocity of the ball - why does this increase with each bounce; the question states that it should increase but I don't understand why and I have to explain why?

c = ? - maximum height? but that does not make sense

8. Nov 4, 2007

### learningphysics

b is not the initial velocity of the ball here... and c isn't the maximum height. because the equation only applies starting with the bounce... (ie it doesn't apply at t = 0)

look at the first fit equation... you need to examine t = 0.8 that's when the bounce takes place...

what is the y value at t = 0.8

what is the velocity at t = 0.8.

Last edited: Nov 4, 2007
9. Nov 4, 2007

### learningphysics

a = -3.783
b = 8.942
c=-4.659

y = -3.783x^2 + 8.942x - 4.659

at x = 0.8

we get y = 0.07348m (just about 0)

dy/dx = -7.566x + 8.942

dy/dx at x = 0.8 is 2.8892m/s

10. Nov 4, 2007

### NKKM

k that makes sense

1) For each interval, write down the values of the three fit parameters A, B and C. By comparing equations for kinematics and the quadratic equation, interpret the meaning of each parameter and how they are related to the position, velocity and acceleration of the ball in the different intervals.

I would have to talk about the graph being shifted- but then how to reconcile those values (a,b,c)?

2)From the fit results of each interval, you may notice that the value of (the fit parameter) B increases as the ball makes a new bounce! If B is interpreted as the initial velocity of the ball for the corresponding bounce, this seems to contradict the observed loss of mechanical energy after each bounce! Can you explain this apparent discrepancy?

for this one the shifting of the graph is again the explanation?

11. Nov 4, 2007

### learningphysics

Yes, the shifting is the reason... think of it like this... the ordinary kinematics starting at x = 0, y = 0 would be:

y = v1*x + (1/2)ax^2

y = v1*x - (1/2)gx^2

now we shift this graph to the right by some value s...

y = v1*(x-s) - (1/2)g(x-s)^2

if you multiply this out then compare to y = ax^2 + bx + c

equate coefficients... that will relate your a, b and c values to v1, g and the time shift s...

you can solve for v1 then from the a, b and c values... and you can check to see if the v1's are decreasing or increasing...

Last edited: Nov 4, 2007
12. Nov 4, 2007

### NKKM

Thanks you so much! I really appreciate your help.... as i work through this I hope there are no more problems :)
thanks

13. Nov 4, 2007

### learningphysics

no prob. :)

14. Feb 22, 2008

### DynamicGirl

hi everyone
I am new to this forum and have come across the same problem. I have a similar graph and a similar function, however I do not follow the discussion above. i am still confused as to how the parameters A B and C relate to the equation y = y0 + v0t - at^2 --- Can you please help to clarify this for me ? I understand that the shift is occuring and that to find y at time = 0.8s you have to substitute that in for t but what exactly do A and B and C mean? I know that A is 0.5 times g. What are B and C???

PPPPLLLLEEEASE help :(

15. Feb 22, 2008

### DynamicGirl

sorry, one more thing..

I can find the velocities for each bounce by plugging in the time that each bounce started however the question specifically states "for each interval write down the vales of the three fit parameters A, B and C. By comparing the equation y = y0 + v0t - at^2 and y = Ax^2 + Bx + C interpret the meaning of each parameter and how they are related to the position, velocity and acceleration of the ball in the different intervals".

Thanks..

16. Feb 22, 2008

### belliott4488

DG - All you're doing is identifying coefficients of a quadratic equation. Does it help to rewrite y(x) as y(t), i.e. just rename the independent variable from x to t, which is what it is in this problem? Then just equate the coefficients of the t^2 term (as you've already done) and the t term, as well as the t^0 term, i.e. the constant. (I assume you know what's meant by y0 and v0.)

Plugging in "special" values of t, like the start of the bounce, can help to identify the meaning of the constants, especially if that is not at t=0, i.e. the equation is shifted in t.

17. Feb 22, 2008

### DynamicGirl

Ok so what I get from what you are you are saying is to do the following:

y0 + v0t - at^2 = At^2 + Bt + C

and equate the coefficients so that B = vo (initial velocity at the beginning of the interval) and and A = a and C = yo?

But as the previous poster pointed out that B is only vo when t = 0. Im not understanding still what B and C correspond to..

LearningPhysics said : "b is not the initial velocity of the ball here... and c isn't the maximum height. because the equation only applies starting with the bounce... (ie it doesn't apply at t = 0)"

so then what are B and C?? theres something im just not getting here..

Last edited: Feb 22, 2008
18. Feb 22, 2008

### belliott4488

When you "plug in" the points from one of those "humps" and ask your calculator to solve for the coefficients, it's fitting a curve right at that location. Therefore, the "initial velocity", i.e. the velocity it has at y=0 will be be the velocity at whatever time corresponds to y=0 for that hump. You could always change variables to, say, t' = t-t0 where t0 is the time when y=0. Then you'd really be starting at time t'=0. That might make it easier to see how the equation relates to the points on the curve.

19. Feb 22, 2008

### DynamicGirl

Thanks for your reply. I got that far. I just am not sure how to answer "interpret the meaning for each parameter A, B and C by comparing the two equations and how they are related to the position velocity and acceleration of the ball at different intervals".

I know that at t = 0 the C is equal to the intial height of the ball, b is equal to initial velocity and a is equal to 1/2 times the acceleration due to gravity. i also know that if i substitute in the time (say of the second hump = 2.15 seconds) then i will be able to solve for y which is the initial height. If i substitute in the time at which the peak occurs i will get the h (y value) at that time. However, the question above is what I am not sure of because on its own:

y = Ax^2 + Bx + C

what does B mean when we are not starting at the bottom of the hump?

do you know what i mean? sorry its so confusing...

20. Feb 22, 2008

### DynamicGirl

heres an example of what i did - my graph is a little different from the one posted by the original poster but close -- on second thought.. i'll just use the numbers of the graph already posted to make it easier..

First interval: occurs at t = 0.8s
The program calculated the following coefficients:
A = -3.783
B = 8.942
C = -4.659

using Y0 = A*X^2 + B*X + C
we get: Y0 = (-3.78)X^2 + (8.942)X + (-4.659)
plugging in the time will give me y0 the initial height of the ball which is 0.0754m

Velocity:
Now i can plug in the time to get y and also to get dy/dt which is the initial velocity:

Y0 = (-3.78)X^2 + (8.942)X + (-4.659)

dy/dx = 2(-3.780)X + 8.942
dy/dx = 2(-3.780)(0.8s) + 8.942
dy/dt = 2.894 m/s

Ok.. so we know how to calculate initial height and velocity but what about the equation
y = Ax^2 + Bx + C

how can we interpret A B and C and relate them to position velocity and acceleration
i already know acceleration is 2a but what do we say about B and C. They are not intiial velocity and position when time is not = 0s so what are they?

21. Feb 22, 2008

### belliott4488

Hi, DG - sorry I keep wandering off ... I hope you haven't been waiting patiently (or worse, impatiently) by your PC. Anyway ...

You've almost got it. You figured out that for the first hump C is the initial position by setting y=0 so that the terms with A and B vanish, and then B is initial velocity by doing the same for the velocity (derivative) equation. For the next hump, though, you don't have t=0 at the start of the hump ... so try shifting t to t' by subtracting the value of t at the start of the hump. That will give you a new equation that gives you initial position and velocity for t'=0. Unfortunately, it will also give you new constants, since subtracting off that t0 value will give you new coefficients after you rearrange things into the form y=A't'^2 + B't' + C' (where I'm using primes for the new values, which will have t0, the start time of the hump, in them). You can now identify your new coefficients as initial position and velocity as you did before. Do that, then rearrange to solve for your original B and C, and you'll get expressions for each of them in terms of the initial position, velocity, and time (what I called t0).

Unfortunately, I don't think there's a simpler interpretation for the second and later humps, since really B and C tell you values of the y-intercepts for the curve and its derivative, but those don't have much physical meaning since each of those curves starts at later time than that.

22. Feb 22, 2008

### learningphysics

It is just as belliott4488 described. There is not much more significance to B and C than that.

If you rewrite the equation given in the form y = At^2 + Bt + C, into the form

y = A'(t-tb)^2 + B'(t-tb) + C', where tb is the time of the bounce being considered... then A' is the acceleration at the time of the bounce at tb... B' is the velocity at the time tb and C' is the displacement at the time tb...

A' = A... but B' does not necessarily equal B, and C' does not necessarily equal C.