Question

A football thrown by a quarterback follows a path given by hx=-0.0095x2+x+7, where h is the height of the ball in feet and x is the horizontal distance the ball has traveled in feet. If any height less than 10 feet can be caught or knocked down, at what distances from the quarterback can the ball be knocked down?

Answer 1

Answer:

The ball can be knocked down at a horizontal distance of 3.09 feet or 102.17 feet from the marshal.

Step-by-step explanation:

We have the function that represents the height h (x) of the ball $h(x) = -0.0095x^2 + x + 7$

Where x is the horizontal distance of the ball.

We want to find the horizontal distance the ball is at (horizontal distance between the field marshal and the ball) when it is at a height of 10 feet.

To do this, we must do h (x) = 10

$10 = -0.0095x^2 + x + 7\\0 = -0.0095x^2 + x -3$

Now we must solve the second degree equation. For this we use the formula of the resolvent:

$\frac{-b + \sqrt{b ^ 2 - 4 * a * c}}{2a}$

       and

$\frac{-b - \sqrt{b ^ 2 - 4 * a * c}}{2a}$

$\frac{-1 + \sqrt{1 ^ 2 - 4 * (- 0.0095)(-3)}} {2 (-0.0095)} = 3.09$ ft

  and

$\frac{-1 - \sqrt{1 ^ 2 - 4 * (- 0.0095)(-3)}} {2 (-0.0095)} = 102.17$ft

Then, the ball can be knocked down at a horizontal distance of 3.09 feet or 102.17 feet from the marshal.