Question

a volley ball player servers the ball. The ball follows a path given by the equation y=-0.01x^2+0.5x+3 where x and y are measured in feet and the origin is on the court directly below where the player hits the ball. for what distance from the player is the ball at least as high as the net, which is 8 ft high? if the player is 30 feet away from the net, is the ball likely to go over the net?

Answer 1

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

$y=-0.01x^2+0.5x+3$

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

$8=-0.01x^2+0.5x+3$

$8\cdot \:100=-0.01x^2\cdot \:100+0.5x\cdot \:100+3\cdot \:100$

$800=-x^2+50x+300$

$-x^2+50x-500=0$

$x^2-50x+500=0$

we can use quadratic formula

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

$x=\frac{-50\pm \sqrt{50^2-4\left(-1\right)\left(-500\right)}}{2\left(-1\right)}$

$x=5\left(5-\sqrt{5}\right),\:x=5\left(5+\sqrt{5}\right)$

$x=13.82,x=36.2$

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

$y=-0.01(30)^2+0.5(30)+3$

$y=9ft$

we know that

height of net is 8 ft

so, the ball will go over the net