Question

Question 617 ptsA model rocket is launched straight up. Its height in feet (y) above theground x seconds after launch is modeled by the quadratic function: y =-16x2 + 208x + 12.How many seconds did it take the rocket to strike the ground?

Answer 1

We have the equation for the height of the rocket:

`y=-16x^2+208x+12`

We have to calculate the time x where the rocket strikes the ground. That is, when y=0.

Then, we have to calculate the roots of the quadratic equation.:

`\begin{gathered} x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ \end{gathered}`
`\begin{gathered} x=\frac{-208\pm\sqrt[]{208^2-4\cdot(-16)\cdot12}}{2\cdot(-16)} \\ x=\frac{-208\pm\sqrt[]{43264+768}}{-32}=\frac{-208\pm\sqrt[]{44032}}{-32}=(-208\pm209.838)/(-32) \\ x_1=(-208+209.838)/(-32)=(1.838)/(-32)=-0.057 \\ x_2=(-208-209.838)/(-32)=(-417.838)/(-32)=13.057 \end{gathered}`

The first solution is like to be the situation of launching, so the valid solution we are looking for is x2=13.057 seconds.