SOLUTION: A ball is thrown across a playing field. Its path is given by the equation below, where x is the distance the ball has traveled horizontally, and y is its height above ground level

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Question 411177: A ball is thrown across a playing field. Its path is given by the equation below, where x is the distance the ball has traveled horizontally, and y is its height above ground level, both measured in feet.
y = -0.005x2 + x + 9
A) What is the maximum height attained by the ball?
B) How far has it traveled horizontally when it hits the ground? Round your answer to the nearest foot.
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
y = -0.005x^2 + x + 9
A) What is the maximum height attained by the ball?
Max occurs when x = -b/2a = -1/(2*-0.005) = 100 seconds
Max height is f(100) = -0.005(100)^2+100+9 = 59
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B) How far has it traveled horizontally when it hits the ground? Round your answer to the nearest foot.
It started at f(0) = 9 ft and will be back on the ground at f(x) = 0 ft.
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Solve -0.005x^2+x+9 = 0
Solve for "x" using the Quadratic Formula.
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Cheers,
Stan H.
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