SOLUTION: A rock is thrown straight up into the air from a height of 4 feet, the height of the rock above the ground , in feet, t seconds after it is thrown is given by -16t^2 + 56T + 4. for
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Question 706424: A rock is thrown straight up into the air from a height of 4 feet, the height of the rock above the ground , in feet, t seconds after it is thrown is given by -16t^2 + 56T + 4. for how many seconds will the height of the rock be at least 28 feet above the ground? Answer by nerdybill(7384) (Show Source):
You can put this solution on YOUR website! h = -16t^2 + 56T + 4
set h to 28 and solve for t:
28 = -16t^2 + 56T + 4
-28 = 16t^2 - 56T - 4
0 = 16t^2 - 56T + 24
0 = 2t^2 - 7T + 3
factoring the left:
0 = (2t-1)(t-3)
t = {.5, 3}
Time in above 28 feet then:
3 - .5 = 2.5 seconds
