SOLUTION: an object is thrown upward from the top of a 48-foot building with an initial velocity of 32 feet per second. the height h of the object after t seconds is given by the quadratic e
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Question 949108: an object is thrown upward from the top of a 48-foot building with an initial velocity of 32 feet per second. the height h of the object after t seconds is given by the quadratic equation h=-16t^2+32t+48. when will the object hit the ground? Answer by rothauserc(4718) (Show Source):
You can put this solution on YOUR website! we are given
h=-16t^2+32t+48
when object hits the ground, h = 0, therefore
0 = -16t^2+32t+48
divide both sides of = by 16
-t^2 +2t + 3 = 0
factor equation
(-t+3)(t+1) = 0
then we have t = 3 and t = -1
we want the positive value for t
the object will hit the ground in 3 seconds
