The height of the ball (in feet) after t seconds is given by the equation:
[tex]h(t)=182-12t-16t^2[/tex]
Note that for t = 0 the height is 182 feet, the initial height. It's required to find the value of t at the moment when the ball hits the ground, that is, when h = 0:
[tex]182-12t-16t^2=0[/tex]
Rearranging:
[tex]-16t^2-12t+182=0[/tex]
This is a quadratic equation with coefficients:
a = -16, b = -12, c = 182.
We need to apply the quadratic formula:
[tex]t=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}[/tex]
Substituting:
[tex]t=\frac{-(-12)\pm\sqrt{(-12)^2-4(-16)(182)}}{2(-16)}[/tex]
Operating:
[tex]t=\frac{12\pm\sqrt{144+11648}}{-32}[/tex]
Calculating:
[tex]t=\frac{12\pm108.591}{-32}[/tex]
There is one positive solution and one negative solution. We only take the positive solution because the time cannot be negative in this context.
The positive solution is:
[tex]t=\frac{12-108.591}{-32}[/tex]
Calculating: t = 3.018 seconds.
Rounding to the nearest hundredth: t = 3.02 seconds
