Question 1102316: If a baseball that is hit follows a parabolic path 72 meters at the base and 16 meters high at the center,find the equation of a parabola that gives the path of the baseball. Note that the departure of the ball is point of origin.
Found 2 solutions by htmentor, Alan3354:
Answer by htmentor(1343)   (Show Source):
You can put this solution on YOUR website! First let's assume the path is symmetric about y-axis.
Then we can shift the parabola along the x-axis to make the departure point the origin.
The equation will take on the form y = ax^2 + c. The vertex will be at the point
(0,16), and the endpoints of the base are (-36,0) and (36,0)
Thus 16 = a*0 + c -> c = 16
Use one of the base points to find a
0 = a*36^2 + 16 -> a = -16/36^2 = -1/81
In order to have the parabola start at the origin, we make a translation along the x-axis:
x -> x-36
The final equation is y = (-1/81)(x-36)^2 + 16
Note that the leading coefficient, a, is equal to -h/(b/2)^2, where h is the height and b is the base
Answer by Alan3354(69443)  (Show Source):
You can put this solution on YOUR website! If a baseball that is hit follows a parabolic path 72 meters at the base and 16 meters high at the center, find the equation of a parabola that gives the path of the baseball. Note that the departure of the ball is point of origin.
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You have 3 points:
(0,0)
(36,16)
(72,0)
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y = ax^2 + bx + c is the parabola
Using the 1st point, c = 0
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From the 2nd point:
16 = a*36^2 + 36b
324a + 9b = 4
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From the 3rd point:
0 = 5184a + 72b
72a + b = 0
b = -72a
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Sub for b in the 2nd point equation:
324a - 648a = 4
a = -1/81
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b = 8/9
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y = -x^2/81 + 8x/9 is the equation
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