Given the function:
[tex]h(t)=-4.9t^2+238t+122[/tex]Where h is the height of the rocket and t is the time in seconds
The rocket launches at t = 0
so, to find the height substitute with t = 0 into the function
so, the height =
[tex]-4.9\cdot0+238\cdot0+122=122[/tex]so, the answer will be when t = 0 , h(t) = 122
How long will the rocket remain in the air ?
So, we will find the value of t when h(t) = 0
[tex]\begin{gathered} -4.9t^2+238t+122=0 \\ a=-4.9,b=238,c=122 \\ \\ t=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}=\frac{-238\pm\sqrt[]{238^2-4\cdot-4.9\cdot122}}{2\cdot-4.9} \\ \\ t=-0.507,\text{ or t = 49.0787} \end{gathered}[/tex]