Answer:
The rocket will hit the ground after about 8.20 seconds.
Step-by-step explanation:
The height of the rocket y, in feet, x seconds after launch is modeled by the equation:
[tex]y=-16x^2+121x+83[/tex]
We want to find the time at which the rocket will hit the ground.
If it hits the ground, the height of the rocket y will be 0. Thus:
[tex]0=-16x^2+121x+83[/tex]
We can solve for x. Factoring (if possible at all) or completing the square can be tedious, so we can use the quadratic formula:
[tex]\displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]
In this case, a = -16, b = 121, and c = 83. Substitute:
[tex]\displaystyle x=\frac{-(121)\pm\sqrt{(121)^2-4(-16)(83)}}{2(-16)}[/tex]
Simplify:
[tex]\displaystyle x=\frac{-121\pm\sqrt{19953}}{-32}[/tex]
Divide everything by -1 and simplify the square root. The plus/minus will remain unchanged:
[tex]\displaystyle x=\frac{121\pm3\sqrt{2217}}{32}[/tex]
Therefore, our two solutions are:
[tex]\displaystyle x=\frac{121+3\sqrt{2217}}{32}\approx 8.20\text{ or } x=\frac{121-3\sqrt{2217}}{32}\approx -0.63[/tex]
Since time cannot be negative, we can ignore the second solution.
Therefore, the rocket will hit the ground after about 8.20 seconds.
