Answer:
The rocket will hit the ground after about 11.63 seconds.
Step-by-step explanation:
A rocket is launched from a tower. Its height y in feet after x seconds is modeled by the equation:
[tex]y=-16x^2+181x+59[/tex]
We want to determine the time at which the rocket will hit the ground.
If it hits the ground, its height y above the ground will be 0. So, we can set y = 0 and solve for x:
[tex]0=-16x^2+181x+59[/tex]
Solve the quadratic. Factoring is too tedious or impossible, and completing the square is also very 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 = 181, and c = 59. Substitute:
[tex]\displaystyle x=\frac{-(181)\pm\sqrt{(181)^2-4(-16)(59)}}{2(-16)}[/tex]
Evaluate:
[tex]\displaystyle x=\frac{-181\pm\sqrt{36537}}{-32}[/tex]
Simplify:
[tex]\displaystyle x=\frac{181\pm\sqrt{36537}}{32}[/tex]
Hence, our solutions are:
[tex]\displaystyle x=\frac{181+\sqrt{36537}}{32}\approx 11.63\text{ or } x=\frac{181-\sqrt{36537}}{32}\approx-0.32[/tex]
Since time cannot be negative, we can ignore the second solution.
So, the rocket will hit the ground after about 11.63 seconds.
