Question 317230
A ball is thrown across a playing field. Its path is given by the equation y=-0.005x^2+x+5. Where x is the distance the ball has traveled horizontally and y is its height above ground level, both measured in feet.
a. What is the maximum height attained by the ball?
The equation:
y=-0.005x^2+x+5
describes a parabola that opens downward (negative coefficient associated with the x^2 term) therefore, finding the vertex will give us the answer.
Axis of symmetry:
x = -b/(2a) = -1/(2*(-0.005)) = 100
Plug the above back into:
y=-0.005x^2+x+5
y=-0.005(100)^2+100+5
y = -50+100+5
y = 55 feet
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b. How far has it traveled horizontally when it hits the ground?
Set y=0 and solve for x:
y=-0.005x^2+x+5
0=-0.005x^2+x+5
Solve using the quadratic formula... doing so yields:
x = {-4.88, 204.88}
We can toss out the negative answer leaving:
x = 204.88 feet
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Details of quadratic to follow:
*[invoke quadratic "x", -0.005, 1, 5 ]

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