SOLUTION: HELP ASAP!!!! An object is propelled vertically upward from the top of a 112-foot building. The quadratic function s(t) = -16t^2 + 176t + 112 models the ball’s height above t
Question 80324: HELP ASAP!!!! An object is propelled vertically upward from the top of a 112-foot building. The quadratic function s(t) = -16t^2 + 176t + 112 models the ball’s height above the ground, s(t) in feet, t seconds after it was thrown. How many seconds does it take until the object finally hits the ground? Round to the nearest tenth of a second if necessary. Found 2 solutions by ankor@dixie-net.com, scott8148:Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! An object is propelled vertically upward from the top of a 112-foot building. The quadratic function s(t) = -16t^2 + 176t + 112 models the ball’s height above the ground, s(t) in feet, t seconds after it was thrown. How many seconds does it take until the object finally hits the ground? Round to the nearest tenth of a second if necessary.
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When the object hit's the ground the height, s(t) = 0 so we have:
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-16t^2 + 176t + 112 = 0
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Fortunately we can simplify things by dividing equation by -16
t^2 - 11t - 7 = 0
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Use the quadratic formula: a=1; b=-11; c=-7 (we only care about the positive solution)
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t = 11.6 sec
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Check solution in original equation:
-16(11.6^2) + 176(11.6) + 112
-2153 + 2041 + 112 = 0
