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madreJ [45]
3 years ago
11

The path of a volley ball thrown over a net is modeled with the function A(x) = -0.02x2 + 0.6x + 5, where x is

Mathematics
1 answer:
LiRa [457]3 years ago
3 0

Answer:

The ball travel horizontally 37 feet before it hits the ground

Step-by-step explanation:

we have

A(x)=-0.02x^2+0.6x+5

where

x ----> is the horizontal distance in feet

A --->  is the altitude of the ball, in feet.

we know that

When the ball hit the ground, the altitude of the ball is equal to zero

so

For A(x)=0

-0.02x^2+0.6x+5=0

solve the quadratic equation

The formula to solve a quadratic equation of the form

ax^{2} +bx+c=0

is equal to

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

in this problem we have

-0.02x^2+0.6x+5=0

so

a=-0.02\\b=0.6\\c=5

substitute in the formula

x=\frac{-0.6\pm\sqrt{0.6^{2}-4(-0.02)(5)}} {2(-0.02)}

x=\frac{-0.6\pm\sqrt{0.76}} {-0.04}

x=\frac{-0.6\pm0.87} {-0.04}

x=\frac{-0.6+0.87} {-0.04}=-6.75

x=\frac{-0.6-0.87} {-0.04}=36.75

therefore

The ball travel horizontally 37 feet before it hits the ground

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