SOLUTION: A ball is launched from the ground straight up into the air at a rate of 80 feet per second. Its height h above the ground (in feet) after t seconds is h = 80t – 16t2. After how ma
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Question 565431: A ball is launched from the ground straight up into the air at a rate of 80 feet per second. Its height h above the ground (in feet) after t seconds is h = 80t – 16t2. After how many seconds is the ball 80 feet high on the way back down? Answer by htmentor(1343) (Show Source):
You can put this solution on YOUR website! A ball is launched from the ground straight up into the air at a rate of 80 feet per second. Its height h above the ground (in feet) after t seconds is h = 80t – 16t2. After how many seconds is the ball 80 feet high on the way back down?
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Substituting h=80 and putting in standard form gives:
-16t^2 + 80t - 80 = 0
Factor:
-16(t^2 - 5t + 5) = 0
t^2 - 5t + 5 = 0
Solve for t using the quadratic formula:
t = (5 +- sqrt(25 - 20))/2
Since we are looking for the time on the way back down, take the larger value:
t = (5 + sqrt(5))/2 = 3.618 seconds
