Respuesta :
Answer:
(a)
Distance from player should be 13.82 feet or 36.2 feet
(b)
The ball will go over the net
Step-by-step explanation:
we are given
The ball follows a path given by the equation
[tex]y=-0.01x^2+0.5x+3[/tex]
where
x and y are measured in feet and the origin is on the court directly below where the player hits the ball
(a)
net height is 8 ft
so, we can set y=8
and then we can solve for x
[tex]8=-0.01x^2+0.5x+3[/tex]
[tex]8\cdot \:100=-0.01x^2\cdot \:100+0.5x\cdot \:100+3\cdot \:100[/tex]
[tex]800=-x^2+50x+300[/tex]
[tex]-x^2+50x-500=0[/tex]
[tex]x^2-50x+500=0[/tex]
we can use quadratic formula
[tex]x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]
[tex]x=\frac{-50\pm \sqrt{50^2-4\left(-1\right)\left(-500\right)}}{2\left(-1\right)}[/tex]
[tex]x=5\left(5-\sqrt{5}\right),\:x=5\left(5+\sqrt{5}\right)[/tex]
[tex]x=13.82,x=36.2[/tex]
So, distance from player should be 13.82 feet or 36.2 feet
(b)
we can plug x=30 and check whether y=8 ft
[tex]y=-0.01(30)^2+0.5(30)+3[/tex]
[tex]y=9ft[/tex]
we know that
height of net is 8 ft
so, the ball will go over the net
