Explanation:
the first equation
x = 3t²
gives us
x/3 = t²
t = sqrt(x/3)
we use this in the second equation
y = -9×sqrt(x/3) + 18
y represents the height of the rocket.
so, to hit the ground height = y = 0.
and we solve
0 = -9×sqrt(x/3) + 18
9×sqrt(x/3) = 18
sqrt(x/3) = 2
x/3 = 4
x = 12
t = sqrt(x/3) = sqrt(12/3) = sqrt(4) = 2
it takes 2 units of time for the rocket to hit the ground.
we could have gotten this by solving the second equation directly for y = 0.
0 = -9t + 18
9t = 18
t = 2
but doing the full thing gives us also the location of hitting the ground : it would be x = 12 units away from the launch location.
so, in regular sentences :
as y stands for the height of the rocket above ground based on the duration (t) of its flight, we find that "hitting the ground" means that that height is 0.
so, we equate the expression in the second equation with 0 and solve for t. that gives us the units of time after the launch when it will hit the ground.