Answer:
The ball lands in about 3.29 seconds.
The given equation is:
[tex]h=-5t^2-3t+64[/tex].
Step-by-step explanation:
I do believe the equation is meant to be:
[tex]h=-5t^2-3t+64[/tex].
We want to find [tex]t[/tex] such that [tex]h=0[/tex] since we want to find the time when the ball has hit the ground.
[tex]0=-5t^2-3t+64[/tex]
Our objective here is to solve this equation using the quadratic formula, [tex]t=\frac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex].
First we need to compare the equation we have to [tex]0=at^2+bt+c[/tex].
[tex]a=-5[/tex]
[tex]b=-3[/tex]
[tex]c=64[/tex]
[tex]t=\frac{-(-3)\pm \sqrt{(-3)^2-4(-5)(64)}}{2(-5)}[/tex]
Simplify the denominator first:
[tex]t=\frac{-(-3)\pm \sqrt{(-3)^2-4(-5)(64)}}{-10}[/tex]
Simplify the inside of the square root:
[tex]t=\frac{-(-3)\pm \sqrt{1289}}{-10}[/tex]
Simplify the -(-3):
[tex]t=\frac{3\pm \sqrt{1289}}{-10}[/tex]
So we have either [tex]t=\frac{3+\sqrt{1289}}{-10}[/tex] or [tex]t=\frac{3-\sqrt{1289}}{-10}[/tex].
Let's both of those into our calculator.
[tex]t=-3.89[/tex] or [tex]t=3.29[/tex]
So the ball lands in about 3.29 seconds.
