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Bouncing Ball Quadratic Graph 
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#1
Nov407, 06:49 PM

P: 8

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 


#2
Nov407, 09:46 PM

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P: 4,125

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
Nov407, 09:49 PM

P: 8

thanks for the tip
http://www.imagevimage.com/images/2_...Ball_Graph.jpg here is the link 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 


#4
Nov407, 10:11 PM

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P: 4,125

Bouncing Ball Quadratic Graph
you won't get the velocity at the point where it bounces by looking at the bvalue... you need to take the derivative of the equation... and plug in the time. 


#5
Nov407, 10:14 PM

P: 8

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
Nov407, 10:15 PM

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P: 4,125

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


#7
Nov407, 10:20 PM

P: 8

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
Nov407, 10:20 PM

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P: 4,125

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. 


#9
Nov407, 10:39 PM

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P: 4,125

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
Nov407, 10:41 PM

P: 8

so to answer this question: 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
Nov407, 10:52 PM

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P: 4,125

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*(xs)  (1/2)g(xs)^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... 


#12
Nov407, 11:21 PM

P: 8

thanks 


#13
Nov407, 11:23 PM

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#14
Feb2208, 09:52 AM

P: 6

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
Feb2208, 01:01 PM

P: 6

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
Feb2208, 01:22 PM

P: 666

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
Feb2208, 01:49 PM

P: 6

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.. 


#18
Feb2208, 04:10 PM

P: 666

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' = tt0 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.



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