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A ball is thrown straight up into the air from an 80-foot high platform with an initial velocity of 64 feet per second. The height of the ball in feet, h, is a function of the time after the ball has been thrown in seconds, t. The approximate height of the ball can be modeled by the quadratic equation \displaystyle h=-16t^2+64t+80h=−16t ^2+64t+80. How long did it take for the ball to hit the ground?
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Answered by
oobleck
It hits the ground when h=0, so just solve
-16t^2+64t+80 = 0
that is, after dividing by -16,
t^2 - 4t - 5 = 0
That factors easily.
-16t^2+64t+80 = 0
that is, after dividing by -16,
t^2 - 4t - 5 = 0
That factors easily.
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