## Algebra and Trigonometry 10th Edition

$x^2 =- 49 (y- 100)$
We are given that $v =28 \ ft/s$ and $s=100 \ ft$. Substitute this in the model: $x^2 =(-\dfrac{v^2}{16}) (y- s)$ $x^2 =[-\dfrac{(28)^2}{16}] (y- 100)$ Our answer is: $x^2 =- 49 (y- 100)$