Question

A bottle rocket is launched straight up. Its height in feet (y) above theground x seconds after launch is modeled by the quadratic function: y =- 16x2 + 116x + 7.How many seconds did it take the rocket to strike the ground?

Answer 1

The eqution of height y in terms of the time x is

$y=-16x^2+116x+7.$

We are asked when the rocket will strike the ground, so that is the same that sayin when will the rocket's height will be zero again, that is, y=0. so we have to solve

$-16x^2+116x+7=0.$

the solutions of this equations are

$x=\frac{-116\pm\sqrt[]{116^2-4(-16)(7)}}{2(-16)}=\mleft\{\begin{aligned}-0.05 \\ 7.3\end{aligned}\mright.$

So, the rocket will strike the ground after 7.3 seconds