## Algebra and Trigonometry 10th Edition

$1107 \ ft$
We are given that $v =28 \ ft/s$ and $s=100 \ ft$ Substitute this in the model: $x^2 =(-\dfrac{v^2}{16}) (y- s)$ $x^2 =[-\dfrac{(198)^2}{16}] (y- 100)$ $x^2 =- 2450.25 (y- 500)$ Plug $y=0$ into the given model to compute the horizontal distance (x). $x^2 =- 2450.25 (0- 500)$ or, $x^2 = 1225125$ or, $x \approx -1107, 1107$ Neglect negative values of distance. So, our answer is: $x =1107 \ ft$