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  1. #1
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    need help with regression problem asap!




    I have to do this problem but I have no idea how to interpret the data! Can somebody pleeeeeeeeeease help me?
    An object is thrown upward from a platform above ground. At various times post release, its height above ground is noted.

    Ignoring air resistance, wind, etc, answer the following:
    How fast was the initial velocity and how high was the platform? The greatest value in the height column is 824. Is this the maximum height the object reached, or did it go higher (explain)? When will it hit ground and how fast will it be going then?
    SUMMARY OUTPUT

    Regression Statistics
    Multiple R 0.397359707
    R Square 0.157894737
    Adjusted R Square -0.010526316
    Standard Error 65.5804849
    Observations 7

    ANOVA
    df SS MS F Significance F
    Regression 1 4032 4032 0.9375 0.377391221
    Residual 5 21504 4300.8
    Total 6 25536

    Coefficients Standard Error t Stat P-value Lower 95% Upper 95% Lower 95.0% Upper 95.0%
    Intercept 712 55.42562584 12.84604349 5.08938E-05 569.5241258 854.4758742 569.5241258 854.4758742
    time 12 12.39354671 0.968245837 0.377391221 -19.85857399 43.85857399 -19.85857399 43.85857399

  2. #2
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    This problem is not modeled appropriately by linear regression - the path of the object will follow a parabola, which is typically of the form y = -(x^2) which is a second-degree polynomial....

    That is why the R square values are so low.....

    You should try to plot the data and determine a best second-degree polynomial fit, and then the first derivative of the equation of that best-fit line should help answer the questions of initial velocity, max height, time of flight, and terminal velocity.

  3. #3
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    Okay I tryed plotting it as a second degree polynomial and this is what I got.

    y= -16x^2+140x+520
    R^2=1

    Can you help figure out the next step?
    Thank you for your help!

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    Take the first derivative of equation (1) to get equation (2)
    (1) y = -16x^2+140x+520 this describes the position or height of the object, y, as a function of time, x

    (2) y' = -32x + 140 this describes the velocity of the object, y' , as a function of time, x

    initial velocity --> set x=0 and solve for y' in equation (2)
    --> initial velocity is when x=0

    max height --> set y'=0 and solve for x in equation (2)
    --> max height is when velocity=0

    time of flight --> set y=0 and solve for x in equation (1) -> you may get 2 answers for x, but only one will make sense

    terminal velocity --> plug the answer for time of flight into x for equation (2) and solve for y'
    --> terminal velocity is when the object has reached the end of flight (hits the ground)

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    as a side note, you do realize that x is the time since launch, right? i'm guessing that that was clear from the original data, but you never know.

    then when reading:

    Quote Originally Posted by JohnM
    max height --> set y'=0 and solve for x in equation (2)
    --> max height is when velocity=0
    i notice john made a typo and forgot to mention that you will need to take the time determined with equation 2 and use that time in equation 1 to obtain the maximum height.

    cheers
    jerry

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    jerry,

    Yes, thanks for catching that miss on max height - I did this one rather quickly...

    good point on x being time after launch - unless you plot the data, it isn't obvious...

    ...but I was trained well and plotted it in Excel before doing anything

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    wow john, you plotted it!

    wish my students would listen when i tell them to make a graph BEFORE they begin their problems.....

    cheers
    jerry

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