SOLUTION: A ball is launched into the sky at 272 feet per second from the roof of a skyscraper 1,344 feet tall. The equation for the ball’s height h at time t seconds is h = -16t2 + 272t + 1

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Question 1116133: A ball is launched into the sky at 272 feet per second from the roof of a skyscraper 1,344 feet tall. The equation for the ball’s height h at time t seconds is h = -16t2 + 272t + 1344. When will the ball strike the ground?
Found 2 solutions by josgarithmetic, Alan3354:
Answer by josgarithmetic(39620) About Me  (Show Source):
You can put this solution on YOUR website!
h=0

-16t%5E2%2B272t%2B1344=0
-
You can simplify that equation.
t%5E2-17t-84=0
%28t-21%29%28t%2B4%29=0

One solution you want and the other you do not want.

Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
A ball is launched into the sky at 272 feet per second from the roof of a skyscraper 1,344 feet tall. The equation for the ball’s height h at time t seconds is h = -16t2 + 272t + 1344. When will the ball strike the ground?
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when h = 0
h = -16t2 + 272t + 1344 = 0
Solve for t.
Ignore the negative value.