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Greeley [361]
2 years ago
6

A teenager who is 5 feet tall throws an object into the air. The quadratic function LaTeX: f\left(x\right)=-16x^2+64x+5f ( x ) =

− 16 x 2 + 64 x + 5 is where f(x) is the height of the object in feet and x is the time in seconds. When will the ball be 10 feet in the air?
Mathematics
1 answer:
tia_tia [17]2 years ago
7 0

Answer:

At approximately x = 0.08 and x = 3.92.

Step-by-step explanation:

The height of the ball is modeled by the function:

f(x)=-16x^2+64x+5

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:

10=-16x^2+64x+5

Subtracting 10 from both sides:

-16x^2+64x-5=0

For simplicity, divide both sides by -1:

16x^2-64x+5=0

We will use the quadratic formula. In this case a = 16, b = -64, and c = 5. Therefore:

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

Substitute:

\displaystyle x=\frac{-(-64)\pm\sqrt{(-64)^2-4(16)(5)}}{2(16)}

Evaluate:

\displaystyle x=\frac{64\pm\sqrt{3776}}{32}

Simplify the square root:

\sqrt{3776}=\sqrt{64\cdot 59}=8\sqrt{59}

Therefore:

\displaystyle x=\frac{64\pm8\sqrt{59}}{32}

Simplify:

\displaystyle x=\frac{8\pm\sqrt{59}}{4}

Approximate:

\displaystyle x=\frac{8+\sqrt{59}}{4}\approx 3.92\text{ and } x=\frac{8-\sqrt{59}}{4}\approx0.08

Therefore, the ball will reach a height of 10 feet at approximately x = 0.08 and x = 3.92.

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