To determine how far from Jackie the ball lands, we need to find the x-intercepts of the quadratic equation representing the path of the football, which is given by:
\[ y = -0.05x^2 + 0.75x + 5 \]
The x-intercepts occur where \( y = 0 \). Therefore, we set the equation to zero:
\[ 0 = -0.05x^2 + 0.75x + 5 \]
To solve for \( x \), we can rearrange the equation:
\[ -0.05x^2 + 0.75x + 5 = 0 \]
To make calculations easier, we can multiply the entire equation by -20 (to eliminate the decimal):
\[ x^2 - 15x - 100 = 0 \]
Now we will use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Here, \( a = 1 \), \( b = -15 \), and \( c = -100 \).
Calculating \( b^2 - 4ac \):
\[ b^2 - 4ac = (-15)^2 - 4(1)(-100) = 225 + 400 = 625 \]
Now plug this into the quadratic formula:
\[ x = \frac{-(-15) \pm \sqrt{625}}{2(1)} = \frac{15 \pm 25}{2} \]
This gives us two possible solutions:
- \( x = \frac{40}{2} = 20 \)
- \( x = \frac{-10}{2} = -5 \) (not meaningful in this context)
Since we are looking for a positive distance from Jackie, the ball lands at \( x = 20 \) feet.
Thus, the answer is:
20 feet