Maybe I'm beating a dead horse here, but I'll give it a go anyway. This is a twostep problem (as mentioned above), and I'll try to break it down step by step. It's a bit longer, but the equations are more familiar, maybe.
Initial givens (for the first ten seconds). This is part 1:
$\displaystyle a=53.9$
From this, we can get:
$\displaystyle v(t)=53.9t$
$\displaystyle y(t)=\frac{1}{2}53.9t^2$
Now plug in $\displaystyle t=10$ seconds:
$\displaystyle y(10)=\frac{1}{2}53.9(10)^2= \boxed{2695(m)}$
$\displaystyle v(10)=53.9*10= \boxed{539(m/s)}$
Now, on to step 2. The givens have changed, we reset our time to 0, because we're already in motion, yadda yadda:
$\displaystyle a=9.8$
$\displaystyle v_0=539$
$\displaystyle y_0=2695$
and can plug these numbers into the equations:
$\displaystyle a(t)=9.8$
$\displaystyle v(t)=at+v_0=9.8t + 539$
$\displaystyle y(t)=\frac{1}{2}at^2+v_0t+y_0=\frac{1}{2}(9.8)t^2+539t+2695$
and we need to solve for v(t)=0, because we want to know how long after the fuel runs out that the velocity hits 0 and the rocket starts to fall and ahem:
$\displaystyle v(t)=0=9.8t+539$
$\displaystyle t=55$
and now we can solve y(55) to get our max height:
$\displaystyle y(55)=\frac{1}{2}(9.8)(55)^2 + 539(55) + 2695$
$\displaystyle y(55) = \boxed{17517.5(m)}$
