Solution:

1. Time taken by each ball:
- First ball: \( t_1 = 5 \, \text{s} \)
- Second ball: \( t_2 = 3 \, \text{s} \) (dropped 2 s later)

2. Heights:
- Height of the first ball:
\[
h_1 = \frac{1}{2} g t_1^2 = \frac{1}{2} \cdot 10 \cdot (5)^2 = 125 \, \text{m}
\]
- Height of the second ball:
\[
h_2 = \frac{1}{2} g t_2^2 = \frac{1}{2} \cdot 10 \cdot (3)^2 = 45 \, \text{m}
\]

3. Difference in heights:
\[
\Delta h = h_1 - h_2 = 125 - 45 = 80 \, \text{m}
\]

 The difference in initial heights is 80 meters.