The formula we need is
[tex]h(t)=-16t^2+v_0 t+h_0[/tex], where v₀ is the starting velocity and h₀ is the initial height. Using the velocity and starting height from our problem we have
[tex]h(t)=-16t^2+160t+0[/tex]. The path of this rocket will be a downward facing parabola, so there will be a maximum. This maximum will be at the vertex of the graph. To find the vertex we start out with [tex]x= \frac{-b}{2a}[/tex], which in our case is [tex]x= \frac{-160}{2(-16)}= \frac{-160}{-32} = 5[/tex]. It will take 5 seconds for the rocket to reach its maximum height. Plugging this back into our formula gives us
[tex]h(5)=-16(5^2)+160(5)+0
\\=-16(25)+800
\\=-400+800=400[/tex]
The rocket's maximum height is 400 feet.
We set our formula equal to zero to find the time it takes to hit the ground, then we factor:
[tex]0=-16t^2+160t+0
\\0=-16t^2+160t
\\0=-16t(t-10)[/tex]
Using the zero product property, we know that either -16t =0 or t-10=0. When -16t=0 is at t=0, when the rocket is launched. t-10=0 gives us an answer of t=10, so the rocket reaches the ground again at 10 seconds.