Respuesta :
Answer:
The ball travel horizontally 37 feet before it hits the ground
Step-by-step explanation:
we have
[tex]A(x)=-0.02x^2+0.6x+5[/tex]
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
[tex]-0.02x^2+0.6x+5=0[/tex]
solve the quadratic equation
The formula to solve a quadratic equation of the form
[tex]ax^{2} +bx+c=0[/tex]
is equal to
[tex]x=\frac{-b\pm\sqrt{b^{2}-4ac}} {2a}[/tex]
in this problem we have
[tex]-0.02x^2+0.6x+5=0[/tex]
so
[tex]a=-0.02\\b=0.6\\c=5[/tex]
substitute in the formula
[tex]x=\frac{-0.6\pm\sqrt{0.6^{2}-4(-0.02)(5)}} {2(-0.02)}[/tex]
[tex]x=\frac{-0.6\pm\sqrt{0.76}} {-0.04}[/tex]
[tex]x=\frac{-0.6\pm0.87} {-0.04}[/tex]
[tex]x=\frac{-0.6+0.87} {-0.04}=-6.75[/tex]
[tex]x=\frac{-0.6-0.87} {-0.04}=36.75[/tex]
therefore
The ball travel horizontally 37 feet before it hits the ground
The ball travel horizontally 37 feet before it hits the ground.
Calculation of the height:
Since
The function should be
[tex]A(x) = -0.02x^2 + 0.6x + 5[/tex]
Here
x is the horizontal distance, in feet, from the starting point
and A is the altitude of the ball, in feet.
Now
[tex]x = \frac{-0.6 \pm \sqrt{06^2 - 4(0002)(5)} }{2(-0.02)}\\\\ x = \frac{-06 \pm \sqrt{76} }{-0.04} \\\\x = \frac{-0.6 \pm 0.87}{-0.04} \\\\x = \frac{0.6 + 0.87}{-0.04} = -6.75 \\\\x = \frac{0.6 - 0.87}{-0.04} = 36.75[/tex]
Hence, The ball travel horizontally 37 feet before it hits the ground.
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