SOLUTION: Would someone be so kind as to help with this?
A ball is thrown over a 5 ft fence. The ball clears the fence, but without much
room to spare. The ball lands 10 ft from the f
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-> SOLUTION: Would someone be so kind as to help with this?
A ball is thrown over a 5 ft fence. The ball clears the fence, but without much
room to spare. The ball lands 10 ft from the f
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Question 523955: Would someone be so kind as to help with this?
A ball is thrown over a 5 ft fence. The ball clears the fence, but without much
room to spare. The ball lands 10 ft from the fence. Using the fence as the
axis of symmetry, write an equation that approximates the path of the ball. (Let the origin be where the fence meets the ground.)
I need an equation in vertex form.
Thanks so much! Answer by Alan3354(69443) (Show Source):
You can put this solution on YOUR website! A ball is thrown over a 5 ft fence. The ball clears the fence, but without much
room to spare. The ball lands 10 ft from the fence. Using the fence as the
axis of symmetry, write an equation that approximates the path of the ball. (Let the origin be where the fence meets the ground.)
I need an equation in vertex form.
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Put the Origin at the vertex.
You have 3 points, (-10,-5), (0,0) & (10,5)
Find the parabola thru the 3 points.
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f(0) = 0, so c = 0
f(-10) = -5 and f(10) = -5
a*100 - 10b = -5
a*100 + 10b = -5
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200a = -10
a = -0.05
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a*100 - 10b = -5
-5 - 10b = -5
b = 0
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To make the Origin the base of the fence,
