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Projectile Motion from table of values.

  1. Feb 21, 2013 #1
    1. The problem statement, all variables and given/known data

    A basketball was shot in the air. Use the following data to produce a horizontal position-time graph and a vertical position-time graph

    1. Determine the initial velocity of the basketball.
    2.What was the maximum height of the basketball.
    3. What is the velocity of the basketball at its lowest point?
    4.Determine the average acceleration. How did you determine this value?


    2. Relevant equations

    aave=Δv/Δt
    v2=v20+2aΔx
    v=v0+at

    3. The attempt at a solution

    I used the initial values of dx and dy as the components of the initial velocity. I found that the initial velocity is 13.2279365m/s at an angle o 89.54*. For the maximum height i was planing to use the initial velocity and acceleration to find the height and then use another equation to find time. However i don't know what the acceleration is and i cannot assume is 9.8m/s^2. Help! Thank you!
     

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    Last edited: Feb 21, 2013
  2. jcsd
  3. Feb 21, 2013 #2

    Simon Bridge

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    Plot the data and sketch the curves in.

    You can work out the acceleration (in the y direction) by plotting the actual dy data and what would have happened had there been no acceleration. Then you'll see how much speed is lost in each time interval. ##a_{ave}=\Delta v/\Delta t##

    Why not use the data to determine the maximum height?

    Aside: be sure to point out that you are estimating the initial velocity from the average velocity in the first 0.18s ... I don't know how accurate you are supposed to be.
     
  4. Feb 22, 2013 #3

    tms

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    Because it is not likely that a measurement was taken exactly when the highest point was reached.
     
  5. Feb 22, 2013 #4

    Simon Bridge

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    Did you get how to find the average acceleration?
    That does not mean you cannot use the data to determine it - it just means the max height is between two measurements. That's why you plot the data and fit a curve ... the curve max will be between two data points ... if the data is good, your eyes/brain are actually quite good at interpolation (or you can do the math - computers are good at this). At the very least it will tell you if the calculated values are plausible.

    The more data you account for in your computation, the better your conclusions will be.
    Conclusions drawn from physical data is always better than conclusions drawn from theoretical calculations.

    For instance: your reasoning seems to assume air resistance can be neglected - is this a good assumption for a basketball? For the particular basketball in your data?
     
  6. Feb 22, 2013 #5

    tms

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    But to accurately get the maximum from the plot, you must actually fit a curve to the data, not just eyeball it. You can't assume that every data point is exactly on the fit curve.

    BTW, I'm not the OP, to whom this last is addressed.
     
  7. Feb 22, 2013 #6

    Simon Bridge

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    Argh: I just failed Reading/Comprehension 101 :)

    OK - I'll reword it: OP's reasoning seems to assume that air resistance is negligible, but this may not be a good assumption for experimental results from a basketball thrown in air in the real world.

    I am hoping to encourage OP to use more of the data when doing the analysis... and to refer back to the data when considering the results of the analysis.
    I can tell from that statement that you have not plotted the data ;)

    Anyway - I know that.
    That is why I commented in post #2 that I did not know how accurate the answer has to be. The "measurements" provided appear to be accurate to sub-atomic scales for example - I suppose it is possible that similar sig fig would be needed in the answer. It is common in HS physics courses, however, for students to fit curves by eye alone. The process does not involve assuming that every data-point lies on the curve.

    When I was designing a junior undergrad physics lab, I wanted to know how good students were at drawing "best fit" lines so I tested about 20 of them. They were well within the uncertainty by linear regression every time. They also did well sketching quadratics... though locating the maxima required more data in that area. Thus that "if the data is good" comment. You should try this some time. [*]

    Whatever - we also know that OP has access to a computer, so may be able to do some sort of regression. We don't know - which is why I asked OP the question. Even if not, plotting the data and sketching in the curve (which, for this data, is feasible) will show if the calculation gives a plausible result.

    i.e. if air resistance is a factor (I've plotted the data - so I know if it is or not) then the calculated value will be obviously wrong compared with the plot.

    Plotting experimental data is usually a good idea anyway.

    ---------------------------

    [*] aside: that's 20 new-entry undergrads I had direct access to ... intregued I tried it with leter years, and found that they got worse at it as they got more advanced.
     
    Last edited: Feb 22, 2013
  8. Feb 22, 2013 #7

    tms

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    Doesn't really matter.
    True.
    I certainly agree with that; I did not mean to suggest otherwise.
    I hadn't when I wrote, but I did afterwards. I suppose the silly number of significant digits should have clued me in.
    I know you do. My comments were more addressed to the OP.
    I did, back when I was an undergrad, and was severely chastised by the instructor for doing so.
    Agreed.
    Interesting.
     
  9. Feb 22, 2013 #8

    Simon Bridge

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    Properly too - the approach does not help you to estimate errors. I mean that you should try comparing the eyeball-best-fit lines with linear regression approaches.

    The person running the lab before me allowed eyeballed lines and I was trying to show it was a bad idea. I wanted students to use the method of crossed lines - which the next-years were doing with error bars.

    Need OP feedback. I think the person who set the problem was being kind :)
     
  10. Feb 22, 2013 #9
    Thank you to both of you. This was immensely helpful and i appreciate the time. I was told not to make assumptions and yet I had to for air resistance.
     
  11. Feb 22, 2013 #10
    I would have to do tangent lines to measure the acceleration in the y-dimension and use that average acceleration to "accurately" find the maxim height?

    And what would be a more accurate way of finding the initial velocity?
     
  12. Feb 22, 2013 #11

    Simon Bridge

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    You don't have to - if you plot the data.

    With no air resistance - what shape is the graph of dy vs t or dy vs dx?
    What shape is the data?

    Estimations from averages are probably fine for this project - it's just good practice to make your reasoning explicit: write down what you do and why. You get more marks that way.

    Note: I would get initial velocity by figuring out what shape the trajectory is and use a regression analysis to fit the general curve to the data. Not so much because it's more accurate but because the method will also tell me how accurate it is. I'm guessing you are not expected to use regression ...
     
  13. Feb 22, 2013 #12
    The shape of dy-time is -x^2, the shape of dx-time is a line. And you are correct we are not expected to use regression.
     
  14. Feb 22, 2013 #13

    Simon Bridge

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    Well - those would be the shapes if the air resistance were negligible.
    Compare with your data. Then you can say whether the data is consistent with zero air resistance and you haven't had to make any assumptions.

    That's the approach each step of the way.
    If you plot the graph, then you can sketch the curve in.
    You must be getting tired of me saying that. These are long answers right? A graph is worth a lot of marks.

    If we have a time series for y, then the average speed in the ith time interval is ##v_i=\Delta y_i/\Delta t_i## and you can do the same with the average acceleration in each time interval - you'll have to decide how many sig fig to keep in the data too.
     
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