If not caught, the football is in the air for about 3.83 seconds
The quadratic function:
[tex]f(x)=-16x^2+60x+5[/tex]
models the height of a football after x seconds, so we want to know how long the ball is in the air if not caught. The ball lands on the ground when [tex]f(x)=0[/tex]. Then, our equation becomes:
[tex]-16x^2+60x+5=0[/tex]
Using quadratic formula we can get the x-values that makes this equation to be true:
[tex]x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a} \\ \\ \\ For: \quad a=-16,\:b=60,\:c=5:\quad \\ \\ x_{1,\:2}=\frac{-60\pm \sqrt{60^2-4\left(-16\right)5}}{2\left(-16\right)}[/tex]
[tex]Two \ solutions: \\ \\ x_{1}=\frac{-60+\sqrt{60^2-4\left(-16\right)5}}{2\left(-16\right)}=-\frac{-15+7\sqrt{5}}{8}=-0.08 \\ \\ \\ x_{2}=\frac{-60+\sqrt{60^2-4\left(-16\right)5}}{2\left(-16\right)}= -\frac{-15+7\sqrt{5}}{8}=3.83[/tex]
Since time can't be negative we discards [tex]x_{1}[/tex] so the only valid solution is [tex]x_{2}=3.83[/tex].
In conclusion: If not caught, the football is in the air for about 3.83 seconds
