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Story Problem w/ parabolas

  1. Nov 4, 2007 #1
    1. The problem statement, all variables and given/known data

    Two children were bouncing a ball. During one particular bounce, the ball traveled a horizontal distance of 6 feet and its maximum height above the ground was 4 feet. The path was in the shape of a parabola. Find the standard equation of the parabola described by this information. Let x represent the horizontal distance in feet from the point the bounce started and y represent the height above the ground in feet.


    Here's the possible answers:

    [​IMG]

    2. Relevant equations

    Standard form of a parabola: F(x)=a(x-h)^2 + k


    3. The attempt at a solution

    Finding a is my problem, but ultimately you need to find the other variables before you can find a so essentially i am having problems with it all.
     
    Last edited: Nov 4, 2007
  2. jcsd
  3. Nov 4, 2007 #2
    In the standard equation k would be be y intercept. In your case I believe that is +4. That narrows it down to A, C, and E. The brute force method is to solve the standard equation when y = 0. You know that when y = 0, that x must equal [tex]\pm[/tex] 6. Therefore you can check equations A, C, and E. to see if when y=0 x=[tex]\pm[/tex]6.
     
  4. Nov 4, 2007 #3
    If its a perfect parabola, then what's the x coordinate of the maximum height?
     
  5. Nov 4, 2007 #4
    I think that the x value for the max height is 3 because it would half way between the 2 x intercepts.

    I had plugged all of the information into the equation, and tried solving for a.

    But i can't come out with any of the a values in any of the possible answers. I get a = 0.
     
  6. Nov 4, 2007 #5

    symbolipoint

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    Starting from standard position, y=a(x-c)^2+b, this has been translated to
    y=a(x-3)^2+4.

    One of the points is (3, 4). Another point on this graph will be (6, 0), one of the intercepts. "a" should be a negative number. Do some algebra and obtain a formula,
    (y-4)=a(x-3)^2
    a = ((y-4)^(0.5))/((x-3)^2)),
    Now plug in the values for coordinate of (6, 0) for x and y and find a. You will obtain the value of a = ? You can figure this out now.
     
  7. Nov 4, 2007 #6
    THANK YOU symbol!

    I figured it out i think, or atleast i got the a value in one of the possible answers.

    Here's how i did it, to help anyone else who might have had a similar problem.

    y=a(x-h)^2 + k (standard equation)

    0=a(6-3)^2 + 4 (In the original problem it says that y represents the height above the ground but the trick is, the height above the ground at a horizontal distance of 6, which means the height is 0 because at a horizontal distance of 6 it has hit the ground already. So i plugged in 3,4 for h,k and i plugged in 6,0 for x,y)

    0=a(3)^2 + 4

    0=9a +4

    -4=9a

    a=-4/9

    Standard equation: y=-4/9(x-3)^2 + 4
     
  8. Nov 5, 2007 #7

    HallsofIvy

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    Actually you don't need to find a at all- the problem does not require you to. You know that the "base" of the parabola goes from 0 to 6 and, because of the symmetry, the vertex will be at the horizontal mid-point, x= 3. The top is at 4 so the vertex is at (3, 4). Now you know that the graph is of the form y= a(x- 3)2+ 40- and only one of the answers, C, is like that.
     
  9. Nov 5, 2007 #8
    Oh yes that is true. But the reason i wanted to make sure and find a was because on the test that's coming up it might not be multiple choice. So i would need to know how to find a in that situation. Thank you so much everyone.
     
  10. Nov 5, 2007 #9

    symbolipoint

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    Absolutely a good approach. I rechecked carefully for "a" to be sure that a correct choice was given among the set of answer choices.
     
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