Answer:
At approximately x = 0.08 and x = 3.92.
Step-by-step explanation:
The height of the ball is modeled by the function:
[tex]f(x)=-16x^2+64x+5[/tex]
Where f(x) is the height after x seconds.
We want to determine the time(s) when the ball is 10 feet in the air.
Therefore, we will set the function equal to 10 and solve for x:
[tex]10=-16x^2+64x+5[/tex]
Subtracting 10 from both sides:
[tex]-16x^2+64x-5=0[/tex]
For simplicity, divide both sides by -1:
[tex]16x^2-64x+5=0[/tex]
We will use the quadratic formula. In this case a = 16, b = -64, and c = 5. Therefore:
[tex]\displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]
Substitute:
[tex]\displaystyle x=\frac{-(-64)\pm\sqrt{(-64)^2-4(16)(5)}}{2(16)}[/tex]
Evaluate:
[tex]\displaystyle x=\frac{64\pm\sqrt{3776}}{32}[/tex]
Simplify the square root:
[tex]\sqrt{3776}=\sqrt{64\cdot 59}=8\sqrt{59}[/tex]
Therefore:
[tex]\displaystyle x=\frac{64\pm8\sqrt{59}}{32}[/tex]
Simplify:
[tex]\displaystyle x=\frac{8\pm\sqrt{59}}{4}[/tex]
Approximate:
[tex]\displaystyle x=\frac{8+\sqrt{59}}{4}\approx 3.92\text{ and } x=\frac{8-\sqrt{59}}{4}\approx0.08[/tex]
Therefore, the ball will reach a height of 10 feet at approximately x = 0.08 and x = 3.92.
