The height of the equation is the sea-level hight, so, whenit splashes into the ocean, h(t) = 0. Let's wirte that down:
[tex]\begin{gathered} h(t)=0 \\ 0=-4.9t^2+235t+318 \end{gathered}[/tex]So, the find t for when the rocket splashes into the ocean, we just have to solve for the roots.
We can use Bhaskara's Formula for that:
[tex]\begin{gathered} a=-4.9 \\ b=235 \\ c=318 \\ t=\frac{-235\pm\sqrt[]{235^2-4\cdot(-4.9)\cdot318}}{2\cdot(-4.9)} \end{gathered}[/tex]Let's calculate, then:
[tex]\begin{gathered} t=\frac{-235\pm\sqrt[]{235^2-4\cdot(-4.9)\cdot318}}{2\cdot(-4.9)}=\frac{-235\pm\sqrt[]{55225+6232.8}}{-9.8} \\ t=\frac{-235\pm\sqrt[]{61457.8}}{-9.8}\approx\frac{-235\pm247.9068}{-9.8} \\ t_1=\frac{-235+247.9068}{-9.8}=\frac{12.9068}{-9.8}\approx-1.3170 \\ t_2=\frac{-235-247.9068}{-9.8}=\frac{-482.9068}{-9.8}\approx49.276 \end{gathered}[/tex]Since the launch was at t = 0, we can't have a negative value of t, so our answer is t = 49.276 s.
To calculate how high above the sea-level the rocket gets, we can firts find the x value of the vertex of the equation, which will be the highest point. This will be our t.
[tex]t_{\max }=x_v=-\frac{b}{2a}=-\frac{235}{2\cdot(-4.9)}=-\frac{235}{-9.8}\approx23.9796[/tex]Now, we input this t into the equation for the height:
[tex]\begin{gathered} h(t_{\max })=-4.9t^2_{\max }+235t_{\max }+318=-4.9\cdot(23.9796)^2+235\cdot23.9796+318 \\ h(t_{\max })=-4.9\cdot575.0212+5635.206+318=5953.206-2817.60388=3135.602 \end{gathered}[/tex]So, the rocket peacks at 3135.602 meters above the sea-level.
