SOLUTION: An object is thrown upward from the top of a 6464​-foot building with an initial velocity of 4848 feet per second. The height h of the object after t seconds is given by the
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-> SOLUTION: An object is thrown upward from the top of a 6464​-foot building with an initial velocity of 4848 feet per second. The height h of the object after t seconds is given by the
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Question 1077480: An object is thrown upward from the top of a 6464-foot building with an initial velocity of 4848 feet per second. The height h of the object after t seconds is given by the quadratic equation h=−16t^2+48t+64. When will the object hit the ground? Answer by Alan3354(69443) (Show Source):
You can put this solution on YOUR website! The height h of the object after t seconds is given by the quadratic equation h=-16t^2 + 48t + 64. When will the object hit the ground?
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When h = 0
-16t^2 + 48t + 64 = 0
Solve for t.
