## Algebra 2 (1st Edition)

The ball will be in the air for about $3.14$ seconds.
Write height model. $h=-16t^{2}+v_{0}t+h_{0} \qquad$ ...substitute $3$ for $h, 50$ for $v_{0},$ and $4$ for $h_{0}$ . $3=-16t^{2}+50t+4\qquad$ ...add $-3$ to each side. $3-3=-16t^{2}+50t+4-3\qquad$...simplify. $0=-16t^{2}+50t+1\qquad$ ...write in standard form. $16t^{2}-50t-1=0\qquad$ ...use the Quadratic formula: $x=\displaystyle \frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ $t=\displaystyle \frac{-(-50)\pm\sqrt{(-50)^{2}-4(16)(-1)}}{2(16)}\qquad$...simplify. $t=\displaystyle \frac{50\pm\sqrt{2564}}{32}\qquad$...simplify. ...since we are calculating time, we can discard the negative solution. $t=\displaystyle \frac{25+\sqrt{641}}{16}\qquad$...use calculator $t\approx 3.14$.