Answer:
y = -1/500x² +2/5x
Explanation:
You want the equation for the path of a football that is thrown 200 m downfield and reaches a maximum height of 20 m.
Initial height
The initial height is not given. The equation is much more easily written if we assume it is zero, or we assume the launch height is the same height at which the ball is caught.
Points
We know the maximum height is reached halfway between the launch point and the final point of interest. Then we're required to write the equation of a parabola that passes through the points (0, 0), (100, 20), and (200, 0).
Equation
Since we know the x-intercepts, we can write the equation as ...
y = ax(x -200)
Then all we have to do is find the value of 'a' so the equation has (100,20) as a solution.
20 = a(100)(100 -200) = -10000a
a = -1/500 . . . . . divide by -10000
The equation of the path of the football is ...
y = (-1/500)(x)(x -200)
y = -1/500x² +2/5x
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Additional comment
When x=185, y = -1/500(185)(185 -200) = 15/500(185) = 5.55 . . . meters
The domain is [0, 200]; the range is [0, 20].
To achieve that distance and height, the football would need to be thrown at a speed in excess of 119 miles per hour. For comparison, the fastest baseball pitch ever thrown was 108.1 miles per hour.