Answer:
The ball is in the air for about 5.873 seconds.
Explanation:
The function:
![f(x)=-5x^2+20x+55](https://img.qammunity.org/2022/formulas/mathematics/high-school/um9bi3v14cx9syc34no8sawecrjc0wsm03.png)
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:
![0=-5x^2+20x+55](https://img.qammunity.org/2022/formulas/mathematics/high-school/1dlgthrsuvrtth6eqohko74z8vfy4fm9qj.png)
We will solve for x. Dividing both sides by -5 yields:
![0=x^2-4x-11](https://img.qammunity.org/2022/formulas/mathematics/high-school/d7pgxfaknytkrzxzdasrd9cc2pybshjtut.png)
The equation is unfactorable, so we can use the quadratic formula:
![\displaystyle x=(-b\pm√(b^2-4ac))/(2a)](https://img.qammunity.org/2022/formulas/mathematics/high-school/iipuedmc7a1mozjrvbx5z80tpb75pjqmzg.png)
In this case, a = 1, b = -4, and c = -11. So:
![\displaystyle x=(-(-4)\pm√((-4)^2-4(1)(-11)))/(2(1))](https://img.qammunity.org/2022/formulas/mathematics/high-school/q4r1rcmhuu0z0il7bsjtngpismuhxyfhdr.png)
Evaluate:
![\displaystyle\begin{aligned} x&=(4\pm√(16+44))/(2)\\&=(4\pm√(60))/(2)\\&=(4\pm2√(15))/(2)\\&=2\pm√(15)\end{aligned}](https://img.qammunity.org/2022/formulas/mathematics/high-school/n38dxfonxvp38cibfk5hr5z57zsbqmhvxr.png)
Approximate:
![x_1=2+√(15)\approx5.873\text{ or } x_2=2-√(15)\approx-1.873](https://img.qammunity.org/2022/formulas/mathematics/high-school/x5paseou85z28adrxdkhrxdrqbiclxin2l.png)
Since time cannot be negative, our only solution is the first choice.
So, the ball is in the air for about 5.873 seconds.