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Math Help - Parabola

  1. #1
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    Parabola

    Hey guys! I'm back with a question I am stuck on... thx for the help!

    A javelin makes a parabola that follows the equation:
    (distance in metres)

    height = -1/25 (distance-15)^2 + 10

    The first question was easy - find the starting height.
    This occurs when d = 0, and you get 1 metre.

    The second question; Find the turning point of the javelin = (15,10)

    But the last question; what is the horizontal distance covered by the javelin has got me stumpted. It contains a square root sign somewhere in the answer I think, but I need some help.

    Thank you!
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  2. #2
    Junior Member hoeltgman's Avatar
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    the horizontal distance is the distance between the starting point and the landing point of the javelin. Just solve the equation and you find two solutions. One can be ecarted because it is a negative one.

    P.S: To solve a second degree use this:

    $ax^2 +bx+c = 0$ \Leftrightarrow x = \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}
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  3. #3
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    But the starting point is 'up' - (0,1), and the finishing point is on the x-axis (the ground), so how do I get a distance by just subbing in the a,b,c values?
    I only get the x-intercept... please help.
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  4. #4
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    Is 30.811 metres correct? I understand now.
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  5. #5
    Junior Member hoeltgman's Avatar
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    It's correct. Precise Value is:
     $15 + 5 \cdot \sqrt{10} $
    Last edited by MathGuru; September 17th 2005 at 10:52 AM.
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