SOLUTION: An Object is thrown upward from the top of a 48-foot building with an initial velocity of 8 feet per second. The height "h" of the object after "t" seconds is given by the quadrati
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Question 1156060: An Object is thrown upward from the top of a 48-foot building with an initial velocity of 8 feet per second. The height "h" of the object after "t" seconds is given by the quadratic equation "h=-16t^2+8t+48" When will the object hit the ground? Found 2 solutions by Alan3354, Boreal:Answer by Alan3354(69443) (Show Source):
You can put this solution on YOUR website! "h=-16t^2+8t+48"
When will the object hit the ground?
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When -16t^2 + 8t + 48 = 0
You can put this solution on YOUR website! h=-16t^2+8t+48=16t^2-8t-48, changing all signs
hits the ground when h=0
factor out an 8
2t^2-t-6=0
t^2-t-12
(t-2)(2t+3)=0
t=2, -3/2, use positiver root of 2 seconds, answer.
