Answer:
The rocket will hit the ground after about 8.20 seconds.
Step-by-step explanation:
The height of the rocket <em>y</em>, in feet, <em>x</em> seconds after launch is modeled by the equation:
![y=-16x^2+121x+83](https://tex.z-dn.net/?f=y%3D-16x%5E2%2B121x%2B83)
We want to find the time at which the rocket will hit the ground.
If it hits the ground, the height of the rocket <em>y</em> will be 0. Thus:
![0=-16x^2+121x+83](https://tex.z-dn.net/?f=0%3D-16x%5E2%2B121x%2B83)
We can solve for <em>x</em>. Factoring (if possible at all) or completing the square can be tedious, so we can use the quadratic formula:
![\displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20x%3D%5Cfrac%7B-b%5Cpm%5Csqrt%7Bb%5E2-4ac%7D%7D%7B2a%7D)
In this case, <em>a</em> = -16, <em>b</em> = 121, and <em>c </em>= 83. Substitute:
![\displaystyle x=\frac{-(121)\pm\sqrt{(121)^2-4(-16)(83)}}{2(-16)}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20x%3D%5Cfrac%7B-%28121%29%5Cpm%5Csqrt%7B%28121%29%5E2-4%28-16%29%2883%29%7D%7D%7B2%28-16%29%7D)
Simplify:
![\displaystyle x=\frac{-121\pm\sqrt{19953}}{-32}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20x%3D%5Cfrac%7B-121%5Cpm%5Csqrt%7B19953%7D%7D%7B-32%7D)
Divide everything by -1 and simplify the square root. The plus/minus will remain unchanged:
![\displaystyle x=\frac{121\pm3\sqrt{2217}}{32}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20x%3D%5Cfrac%7B121%5Cpm3%5Csqrt%7B2217%7D%7D%7B32%7D)
Therefore, our two solutions are:
![\displaystyle x=\frac{121+3\sqrt{2217}}{32}\approx 8.20\text{ or } x=\frac{121-3\sqrt{2217}}{32}\approx -0.63](https://tex.z-dn.net/?f=%5Cdisplaystyle%20x%3D%5Cfrac%7B121%2B3%5Csqrt%7B2217%7D%7D%7B32%7D%5Capprox%208.20%5Ctext%7B%20or%20%7D%20x%3D%5Cfrac%7B121-3%5Csqrt%7B2217%7D%7D%7B32%7D%5Capprox%20-0.63)
Since time cannot be negative, we can ignore the second solution.
Therefore, the rocket will hit the ground after about 8.20 seconds.