SOLUTION: The shell fired from the base of the hill follows the parabolic path y=-(x^2/4)+8x with distances in miles. the hill has a slope of 1/4. how far from the gun is the point of impact

Algebra ->  Coordinate-system -> SOLUTION: The shell fired from the base of the hill follows the parabolic path y=-(x^2/4)+8x with distances in miles. the hill has a slope of 1/4. how far from the gun is the point of impact      Log On


   

Question 636919: The shell fired from the base of the hill follows the parabolic path y=-(x^2/4)+8x with distances in miles. the hill has a slope of 1/4. how far from the gun is the point of impact?
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
The shell fired from the base of the hill follows the parabolic path y=-(x^2/4)+8x with distances in miles. the hill has a slope of 1/4. how far from the gun is the point of impact?
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h(x) = (-1/4)x^2 + 8x is path of the shell where x is in seconds.
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y = (1/4)x is the equation of the slope of the hill.
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Find the intersection of the shell path and the hill:
(1/4)x = (-1/4)x^2+8x
(-1/4)x^2 + (31/4)x = 0
Factor:
(-1/4)x[x - 31) = 0
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Positive solution:
x = 31
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To figure out the slant distance sketch the shell path and the hill
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You have a right triangle with base = 31, height = h(31) = 7.75
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The slant distance = sqrt[31^2+7.75^2] = 31.95
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Cheers,
Stan H.