Step 1: Efficiency of a machine is defined as
\[
\text{Efficiency} = \frac{\text{Useful energy output}}{\text{Energy input}}.
\]
Step 2: Given:
\[
\text{Efficiency} = 90% = 0.9,
\quad \text{Energy input} = 5000\,\text{J}.
\]
Useful energy gained by the mass (increase in potential energy):
\[
PE = 0.9 \times 5000 = 4500\,\text{J}.
\]
Step 3: When the mass is released, this potential energy is completely converted into kinetic energy just before hitting the ground (neglecting losses).
\[
KE = PE = 4500\,\text{J}.
\]
Hence, the kinetic energy on hitting the ground is
\[
\boxed{4500.00\,\text{J}}.
\]