The height in meters is given as a function of t (time) as;
[tex]h(t)=-4.9t^2+268t+416[/tex]
First, let's solve the quadratic function using quadratic formula given as;
[tex]\begin{gathered} t=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ \text{Where a=-4.9} \\ b=268\text{ and c=416} \end{gathered}[/tex][tex]\begin{gathered} t=\frac{-268\pm\sqrt[]{268^2-4(-4.9)(416)}}{2(-4.9)} \\ t=\frac{-268\pm\sqrt[]{71824+8153.6}}{-9.8} \\ t=\frac{-268\pm282.80}{-9.8} \\ t=\frac{-268+282.80}{-9.8}\text{ or t=}\frac{-268-282.80}{-9.8} \\ t=-1.51\text{ or 56.20} \end{gathered}[/tex]
Since a time cannot be a negative value. Hence, the rocket splashes down after 56.20seconds.
Also, to find the peak, let's find the time
[tex]\begin{gathered} h(t)=-4.9t^2+268t+416 \\ h^{\prime}(t)=-9.8t+268 \\ At\text{ the peak, h'(t)=0} \\ \text{That is;} \\ -9.8t+268=0 \\ t=\frac{268}{9.8} \\ t=27.35\text{seconds} \end{gathered}[/tex]
Thus the h(t) above the sea level is at time t=27.35seconds is;
[tex]\begin{gathered} h(27.35)=-4.9(27.35)^2+268(27.35)+416 \\ h(27.35)=4080.49m \end{gathered}[/tex]
The rocket peaks at 4080.49meters above the sea level
