Answer:
The ball hits the ground at approximately 3.25 seconds.
Step-by-step explanation:
It hits the ground when the distance between the ball and the ground is 0. In other words, when h is 0.
We need to solve the following equation:
[tex]0=205-11t-16t^2[/tex]
Compare this to [tex]ax^2+bx+c=0[/tex] and you should determine the values for [tex]a,b,\text{ and } c[/tex].
[tex]a=-16[/tex]
[tex]b=-11[/tex]
[tex]c=205[/tex].
Plug these values into the quadratic formula:
[tex]x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]
[tex]x=\frac{-(-11)\pm \sqrt{(-11)^2-4(-16)(205)}}{2(-16)}[/tex]
[tex]x=\frac{11\pm \sqrt{121+13120}}{-32}[/tex]
[tex]x=\frac{11\pm \sqrt{13241}}{-32}[/tex]
[tex]x=\frac{11+\sqrt{13241}}{-32} \text{ or } \frac{11-\sqrt{13241}}{-32}[/tex]
Let's type both into a calculator:
[tex]x \approx -3.93967 \text{ or } 3.25217[/tex]
So the ball hits the ground at approximately 3.25 seconds.
