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A rock is thrown from a roof and follows a parabolic through the air. The path of the rock is modeled by h(t)=-t^2 +9t+2, where h gives the height of the rock after t seconds since its release. when does the rock first reach a hight higher than 22 meters? for how long does it stay above 22 meters?.

 Apr 25, 2022

Best Answer 

 #1
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\(22=-t^2+9t+2\)

\(t^2 - 9t + 20 = 0\)

\(-4 \cdot -5 = 20, -4-5=-9\)

\((t-4)(t-5) = 0\)

 

\(t - 4 = 0\)

\(t = 4\)

 

\(t - 5 = 0\)

\(t = 5\)

 

The rock will reach a height higher than 22 meters \(\boxed{4}\) seconds after it's released. It will stay above 22 meters for \(5 - 4 = \boxed{1}\) seconds.

 Apr 25, 2022
 #1
avatar
+1
Best Answer

\(22=-t^2+9t+2\)

\(t^2 - 9t + 20 = 0\)

\(-4 \cdot -5 = 20, -4-5=-9\)

\((t-4)(t-5) = 0\)

 

\(t - 4 = 0\)

\(t = 4\)

 

\(t - 5 = 0\)

\(t = 5\)

 

The rock will reach a height higher than 22 meters \(\boxed{4}\) seconds after it's released. It will stay above 22 meters for \(5 - 4 = \boxed{1}\) seconds.

Guest Apr 25, 2022

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