The equation for the height of the rocket at time t given
![h= -16t^2+192t](https://tex.z-dn.net/?f=%20h%3D%20-16t%5E2%2B192t%20)
We have to find the time t, when the rocket reaches 560 feet.
That means we have to find t when h = 560 ft. we will place 560 in the place of h to find t now.
![h= -16t^2+192t](https://tex.z-dn.net/?f=%20h%3D%20-16t%5E2%2B192t%20)
![560 = -16t^2+192t](https://tex.z-dn.net/?f=%20560%20%3D%20-16t%5E2%2B192t%20)
In the right side, we can check -16 is the common factor. So we will take out -16 from the rigbht side.
![560 = -16(t^2 - 12t)](https://tex.z-dn.net/?f=%20560%20%3D%20-16%28t%5E2%20-%2012t%29%20)
To get rid of -16 from the right side and move it to left side, we will divide both sides by -16.
![560/-16 = -16(t^2-12t)/-16](https://tex.z-dn.net/?f=%20560%2F-16%20%3D%20-16%28t%5E2-12t%29%2F-16%20)
![-35 = t^2 -12t](https://tex.z-dn.net/?f=%20-35%20%3D%20t%5E2%20-12t%20)
Now we will move -35 to the righ side by adding 35 to both sides.
![-35+35 = t^2-12t+35](https://tex.z-dn.net/?f=%20-35%2B35%20%3D%20t%5E2-12t%2B35%20)
![0 = t^2 -12t+35](https://tex.z-dn.net/?f=%200%20%3D%20t%5E2%20-12t%2B35%20)
![t^2-12t+35 = 0](https://tex.z-dn.net/?f=%20t%5E2-12t%2B35%20%3D%200%20)
We will factorize thee left side to find the values of t now. We need to find a pair of factors of 35 that by adding them we will get -12.
The pair of factors of 35 are -5 and -7 and by adding -5-7 we will get -12.
![t^2-12t+35 =0](https://tex.z-dn.net/?f=%20t%5E2-12t%2B35%20%3D0%20)
![(t-5)(t-7) =0](https://tex.z-dn.net/?f=%20%28t-5%29%28t-7%29%20%3D0%20)
So by using zero product property we will get
![t-5 =0](https://tex.z-dn.net/?f=%20t-5%20%3D0%20)
![t-5+5 = 0+5](https://tex.z-dn.net/?f=%20t-5%2B5%20%3D%200%2B5%20)
![t=5](https://tex.z-dn.net/?f=%20t%3D5%20)
Also ![t-7 =0](https://tex.z-dn.net/?f=%20t-7%20%3D0%20)
![t-7+7 = 0+7](https://tex.z-dn.net/?f=%20t-7%2B7%20%3D%200%2B7%20)
![t=7](https://tex.z-dn.net/?f=%20t%3D7%20)
So we have got the rocket reaches at 560ft when t = 5 seconds and also when t = 7 seconds.
Now part b.
When the rocket completes its trajectory and hits the ground then the height or h = 0. So we will place h = 0 there in the equation.
![h= -16t^2+192t](https://tex.z-dn.net/?f=%20h%3D%20-16t%5E2%2B192t%20)
![0= -16t^2 + 192 t](https://tex.z-dn.net/?f=%200%3D%20-16t%5E2%20%2B%20192%20t%20)
![0 = -16(t^2-12t)](https://tex.z-dn.net/?f=%200%20%3D%20-16%28t%5E2-12t%29%20)
![-16(t^2-12t) = 0](https://tex.z-dn.net/?f=%20-16%28t%5E2-12t%29%20%3D%200%20)
We will move -16 to the other side by dividing it to both sides.
![-16(t^2-12t)/-16 = 0/-16](https://tex.z-dn.net/?f=%20-16%28t%5E2-12t%29%2F-16%20%3D%200%2F-16%20)
![t^2-12t = 0](https://tex.z-dn.net/?f=%20t%5E2-12t%20%3D%200%20)
We will take out the common factor t from the left side. By taking out t we will get,
![t(t-12) = 0](https://tex.z-dn.net/?f=%20t%28t-12%29%20%3D%200%20)
We will use zero product property now. By using that we will get,
![t = 0](https://tex.z-dn.net/?f=%20t%20%3D%200%20)
ans also ![t-12 = 0](https://tex.z-dn.net/?f=%20t-12%20%3D%200%20)
![t-12+12 = 0+12](https://tex.z-dn.net/?f=%20t-12%2B12%20%3D%200%2B12%20)
![t = 12](https://tex.z-dn.net/?f=%20t%20%3D%2012%20)
When the rocket completes its trajectory and hits the ground the time t can not be 0. When t =0, the rocket starts the trajectory.
So when the rocket completes its trajectory and hits the ground ,
then t = 12seconds.
So we have got the required answers.