SOLUTION: an object is dropped from 45 ft below the tip of the pinnacle atop a 945 ft tall building. the height h of the object after t seconds is given by the equations h= -16t^2+900. find

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Question 296284: an object is dropped from 45 ft below the tip of the pinnacle atop a 945 ft tall building. the height h of the object after t seconds is given by the equations h= -16t^2+900. find how many seconds pass before the object reaches the ground.
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
an object is dropped from 45 ft below the tip of the pinnacle atop a 945 ft tall building. the height h of the object after t seconds is given by the equations h= -16t^2+900. find how many seconds pass before the object reaches the ground.
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The height is zero when the object hits the ground.
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Solve: -16t^2+900 = 0
16^2 = 900
t^2 = 900/16
Positive solution:
t = 30/4 = 15/2 = 7.5 seconds
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Cheers,
Stan H.
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