SOLUTION: An arrow is shot directly upward from the top of a 112-foot tall building with an initial velocity of 96 feet per second. The height of the arrow above the ground after (t) seconds

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Question 288957: An arrow is shot directly upward from the top of a 112-foot tall building with an initial velocity of 96 feet per second. The height of the arrow above the ground after (t) seconds is given by the position function. s(t)=-16t^2+96t+112. After how many seconds will the arrow strike the ground?
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
An arrow is shot directly upward from the top of a 112-foot tall building with an initial velocity of 96 feet per second. The height of the arrow above the ground after (t) seconds is given by the position function.
s(t)=-16t^2+96t+112. After how many seconds will the arrow strike the ground?
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Solve -16t^2 + 96t + 112 = 0
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-16(t^2 - 6t + 7) = 0
Factor:
(t-6)(t+1) = 0
Positive solution:
t = 6 seconds
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Cheers,
Stan H.