Answer:
The ball is in the air for about 5.873 seconds.
Step-by-step explanation:
The function:
[tex]f(x)=-5x^2+20x+55[/tex]
Models the height of a ball x seconds after it is thrown in the air.
And we want to find the total time the ball is in the air.
So, we can simply find the time x at which the ball lands. If it lands, its height f above the ground will be 0. Thus:
[tex]0=-5x^2+20x+55[/tex]
We will solve for x. Dividing both sides by -5 yields:
[tex]0=x^2-4x-11[/tex]
The equation is unfactorable, so we can use the quadratic formula:
[tex]\displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]
In this case, a = 1, b = -4, and c = -11. So:
[tex]\displaystyle x=\frac{-(-4)\pm\sqrt{(-4)^2-4(1)(-11)}}{2(1)}[/tex]
Evaluate:
[tex]\displaystyle\begin{aligned} x&=\frac{4\pm\sqrt{16+44}}{2}\\&=\frac{4\pm\sqrt{60}}{2}\\&=\frac{4\pm2\sqrt{15}}{2}\\&=2\pm\sqrt{15}\end{aligned}[/tex]
Approximate:
[tex]x_1=2+\sqrt{15}\approx5.873\text{ or } x_2=2-\sqrt{15}\approx-1.873[/tex]
Since time cannot be negative, our only solution is the first choice.
So, the ball is in the air for about 5.873 seconds.
