Solution:
Distance travelled by A S=½ gt2 ( u=0 because free fall)
Distance travelled by B S'=½ g(t-2)2 ( as B is dropped after 2 sec)
Difference in position 100 m given
S-S'=100 m
Thus ½ gt2 - ½ g(t-2)2=100
take g=10
t2 - (t-2)2=20
4t - 4=20
t=6 sec
