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?.

Guest Nov 15, 2020

#1**+1 **

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 = -t^{2} + 9t + 2

Subtract 22 from both sides of the equation

0 = -t^{2} +9t - 20

Multiply both sides of the equation by -1

0 = t^{2} - 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

hectictar Nov 15, 2020