As the body is dropped from a certain height h, its initial velocity will be zero.
For first half, h1=2h.
Using second equation of motion, s=ut+21at2
We have, 2h=21gt12.
Time taken in travel first half of the height, t1=gh...(1).
For total height h, using again second equation of motion,
h=21g(t1+t2)2⇒(t1+t2)=g2h...(2)
From equation (1) and (2),
⇒21=t1+t2t1
⇒1+t1t2=2
⇒t2t1=2−11
Thus, the relation is t2=(2−1)t1.