Step 1: Use conservation of mechanical energy.
The cannon ball is fired from a height of
\[
h=55\,\text{m}
\]
with initial speed
\[
u=50\,\text{m s}^{-1}.
\]
When it reaches the ground, its gravitational potential energy is converted into additional kinetic energy.
Step 2: Write the energy relation.
Using
\[
v^2=u^2+2gh,
\]
where \(v\) is the speed while hitting the ground.
Substituting the values,
\[
v^2=50^2+2(10)(55)
\]
\[
v^2=2500+1100
\]
\[
v^2=3600
\]
Step 3: Find \(v\).
\[
v=\sqrt{3600}
\]
\[
v=60\,\text{m s}^{-1}
\]
Step 4: Final conclusion.
Therefore, the speed of the cannon ball while hitting the ground is
\[
\boxed{60\,\text{m s}^{-1}}
\]