1) Let's find the maximum point of this quadratic equation, by using the following formula to get the h coordinate of the vertex and then plug into the function the value we found.
[tex]\begin{gathered} y=-16t^2+80t+2400 \\ h=-\frac{b}{2a}=\frac{-80}{2(-16)}=\frac{5}{2}=2.5 \\ k=-16(\frac{5}{2})^2+80(\frac{5}{2})+2400 \\ k=-16\cdot\frac{25}{4}+200+2400 \\ k=2500 \end{gathered}[/tex]So the maximum height is given y the point at the vertex (2.5,2500) .i.e. 2500 feet
2) From the previous part, we can tell It takes 12.5 seconds for the rocket to get 2500 ft high.
[tex]15-(-10)=\frac{25}{2}=12.5[/tex]We are considering the rocket to strt flying off at x=-10.
3) To find when the rocket reaches the ground is to find the roots of this quadratic equation, and then pick one of them. So let's do it:
[tex]\begin{gathered} -16t^2+80t+2400\:=0 \\ t=\frac{-80\pm\sqrt{80^2-4\left(-16\right)\cdot\:2400}}{2\left(-16\right)} \\ t_1=\frac{-80+400}{2\left(-16\right)}=-10 \\ t_2=\frac{-80-400}{2(-16)}=15 \end{gathered}[/tex]Since we cannot deal with the negative measurement of time. We can tell that
Note that the question is after it is fired consider the rocket is fired at x=-15 and hits its peak 2.5 seconds so the rocket hits the ground 25 seconds after it was launched at t=1
