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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?.

 Nov 15, 2020
 #1
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We want to find the value of  t  when the height is  22.  So let's plug in  22  for  h  and solve for  t .

 

22  =  -t2 + 9t + 2

                                   Subtract  22  from both sides of the equation

0  =  -t2 +9t - 20

                                   Multiply both sides of the equation by  -1

0  =  t2 - 9t + 20

                                   Now we want to factor the right side. What two numbers add to  -9  and multiply to  20 ?  -4  and  -5

0  =  (t - 4)(t - 5)

                                   Set each factor equal to zero and solve for  t

 

t - 4 = 0      or      t - 5 = 0

 

 t  =  4                   t = 5

 

The rock first reaches  22  meters at time 4 seconds.

 

Then it goes above 22 meters and comes back down to 22 meters at time 5 seconds.

 

So it stays above 22 meters for  5 seconds - 4 seconds  =  1 second

 

Here's a graph:  https://www.desmos.com/calculator/zvmpzfjujr

 Nov 15, 2020

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