Question

A rocket is launched into the air and follows the path, h (t) = -3 * t^2 + 12 * t

t is time measured in seconds. Determine how long it takes the rocket to hit the ground. Show all work for full credit.

My lesson didn't really explain this well for me. Either that or I'm not getting it. Could someone explain this a bit better? Thanks!

Answer 1

check the picture below.

so, the rocket will come back to the ground when h(t) = 0, thus

$\bf h(t)=-3t^2+12t\implies \stackrel{h(t)}{0}=-3t^2+12t\implies 0=-3t(t-4)\\\\[-0.35em]~\dotfill\\\\0=-3t\implies 0=t\impliedby \textit{0 seconds when it took off from the ground}\\\\[-0.35em]~\dotfill\\\\0=t-4\implies 4=t\impliedby \textit{4 seconds later, it came back down}$