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?.
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