
#1
Mar2710, 03:27 PM

P: 6

1. The problem statement, all variables and given/known data
In the experiment that I have performed, the ball was thrown upward and caught once it reached its initial position. The data was recorded using the graphical analysis program. The graph produced is a parabola. The parabolic curve was generated showing the values of three parameters: A:4.453, B:9.835, C:3.565. I need to: 1). 1.Interpret the meaning of the fit coefficients A, B and C. 2). Find out the value of the acceleration due to gravity predicted by the experiment. Compare this to the expected value, and calculate the percentage difference. 3). Calculate the quantity −B/(2A), including units, and give a physical meaning to this quantity? I am stuck, as I have not touched the physics book for more than 10 years. Any help will be appreciated. 2. Relevant equations Y = A t2 + B t + C 3. The attempt at a solution 1). The coefficient A is acceleration, B  velocity, C initial position of the ball? 



#2
Mar2710, 05:04 PM

HW Helper
P: 6,210

Yes that is correct.
For the second part, do you know a kinematic equation that resembles your formula of Y=At^{2}+Bt+C? For the third part, what do the units of B/A give? 



#3
Mar2710, 05:42 PM

P: 6

Thank you rock.freak, 2). The kinematic formula is x=x0+v0t+1/2at^2 3). I know that b/a would give me meters, and I assume by dividing B/2A, I would get the highest point (the highest position the ball reaches). But when I look at the graph, it does not make sense to me. By dividing B/2A I get 1.1m, while the graph shows the highest point of approx. 1.8m. Hope you can see the image I attached. 



#4
Mar2710, 08:51 PM

HW Helper
P: 6,210

Motion with Uniform AccelerationHow exactly did you get the first set of readings? 



#5
Mar2710, 09:33 PM

P: 6





#6
Mar2810, 12:07 AM

HW Helper
P: 6,210





#7
Mar2810, 12:18 AM

P: 6

I appreciate your help. I am still stuck at part two. You refered to the kinematick equation x=x0+v0t+1/2at^2, is it to calculate the acceleration of the ball in motion? if yes, when I plug in the values for time and initial velocity, I get a 2.43 for acceleration. And that does not make sense to me. I appreciate your patience, if you have any other suggestion. 



#8
Mar2810, 12:28 AM

HW Helper
P: 6,210

Thus similarly, if the coefficient of t^{2} in the kinematic equation is acceleration, then shouldn't they be the same? i.e. 1/2g = A ? (negative since upwards is taken as the positive direction) 


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