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lolz55678

10.02.2022

Mathematics

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

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

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

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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