Question

A ball is thrown from a height of 205 feet with an initial downward velocity of 11 ft/s. The ball's height h (in feet) after t seconds is given by the following.
REST IS ATTACHED BELOW!!!!!!!! HELP QUICK

Answer 1

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:

$0=205-11t-16t^2$

Compare this to $ax^2+bx+c=0$ and you should determine the values for $a,b,\text{ and } c$.

$a=-16$

$b=-11$

$c=205$.

Plug these values into the quadratic formula:

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$x=\frac{-(-11)\pm \sqrt{(-11)^2-4(-16)(205)}}{2(-16)}$

$x=\frac{11\pm \sqrt{121+13120}}{-32}$

$x=\frac{11\pm \sqrt{13241}}{-32}$

$x=\frac{11+\sqrt{13241}}{-32} \text{ or } \frac{11-\sqrt{13241}}{-32}$

Let's type both into a calculator:

$x \approx -3.93967 \text{ or } 3.25217$

So the ball hits the ground at approximately 3.25 seconds.