Question 87299: a ball is thrown into the air in such a way that its distance (d), in feet, above the ground at any time (t), in seconds, is given by the equation d=40t-5t^2. how many seconds will it take for the ball to reach a hight of 80ft above the ground?
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! a ball is thrown into the air in such a way that its distance (d), in feet, above the ground at any time (t), in seconds, is given by the equation d=40t-5t^2. how many seconds will it take for the ball to reach a hight of 80ft above the ground?
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40t - 5t^2 = d
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Substitute 80 for d and solve for t:
40t - 5t^2 = 80
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Arrange as a quadratic equation:
-5t^2 + 40t - 80 = 0
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Simplify and change the signs, divide equation by -5, and you have:
+1t^2 - 8t + 16 = 0
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Factors easily to:
(t-4)(t-4) = 0
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t = 4 seconds to reach 80 ft
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Check solution by substituting 4 for t in the original equation, find d:
d = 40t - 5t^2
d = 40(4) - 5(4^2)
d = 160 - 5(16)
d = 160 - 80
d = 80 ft
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Did this make sense to you? Not that hard, right?
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