Okay, this is a physics question, but whatever.
Given the initial height, let's call it \(\large y_0\), 110 ft; We are looking for the final height, \(\large y\). We also know that when the rocket will reach its highest point, it's velocity is 0, \(\large v=0\), and the acceleration would be \(\large -32ft/s^2\).
From the free-fall equations of kinematics, using the formula:
\(\large v^2=v_0^2+2a(y-y_0)\)
So,
\(\large 0=(244ft/s)^2-2a(y-110ft)\\
\large 59536=2a(y-110)\\\large
930.25=y-110\\\large y=930.25+110=1040.25\)
The final height would be 1040.25 feet in the sky.