Using quadratic function concepts, it is found that:
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The height of the ball after x seconds is given by:
[tex]y = -\frac{16}{2025}x^2 + \frac{9}{5}x + 1.5 = -0.0079x^2 + 1.8x + 1.5[/tex]
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The height when it was punted is the initial height, that is, y(0).
[tex]y(0) = -0.0079(0)^2 + 1.8(0) + 1.5 = 1.5[/tex]
The ball was 1.5 feet high when it was punted.
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The maximum height is the value of y at the vertex, given by:
[tex]y_V = -\frac{\Delta}{4a}[/tex]
In which:
[tex]\Delta = b^2 - 4ac = (1.8)^2 - 4(-0.0079)(1.5) = 3.2874[/tex]
[tex]y_V = -\frac{3.2874}{4(-0.0079)} = 104[/tex]
The maximum height is of 104 feet.
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The length of the punt is the vertical distance x when y = 0, which we find solving the quadratic equation. Thus:
[tex]x_1 = \frac{-b + \sqrt{\Delta}}{2a} = \frac{-1.8 + \sqrt{3.2874}}{2(-0.0079)} = -0.83[/tex]
[tex]x_2 = \frac{-b - \sqrt{\Delta}}{2a} = \frac{-1.8 - \sqrt{3.2874}}{2(-0.0079)} = 283.7[/tex]
The punt was of 283.7 ft long.
A similar problem is given at https://brainly.com/question/3243250
