Respuesta :
check the picture below. That's about what a parabola for an "initial velocity" case would look like.
anyhow, it hits the ground at two points, when it began, and when it came back down, and that happens when y = 0.
[tex]\bf h(x)=-2x^2+8x\implies \stackrel{h(x)}{0}=-2x^2+8x\implies 0=-2x(x-4)\\\\ -------------------------------\\\\ 0=-2x\implies 0=x\qquad \impliedby \textit{when it began}\\\\ -------------------------------\\\\ 0=x-4\implies \boxed{4=x}\qquad \impliedby \textit{when it came back down}[/tex]
anyhow, it hits the ground at two points, when it began, and when it came back down, and that happens when y = 0.
[tex]\bf h(x)=-2x^2+8x\implies \stackrel{h(x)}{0}=-2x^2+8x\implies 0=-2x(x-4)\\\\ -------------------------------\\\\ 0=-2x\implies 0=x\qquad \impliedby \textit{when it began}\\\\ -------------------------------\\\\ 0=x-4\implies \boxed{4=x}\qquad \impliedby \textit{when it came back down}[/tex]

Answer:
4 seconds
Step-by-step explanation:
h(t) = -3t^2 + 12t
0 = -3t^2 + 12t
0 = -3t(t - 4)
0 = t - 4
t = 4
Therefore, it takes 4 seconds for the rocket to hit the ground.
hope it helps :)
mark brainliest!