Given: A ball is released from rest from height $h$.

So motion is under gravity with $u=0, \quad a=g$

Distance fallen in time $t$ is $s=\frac{1}{2}gt^2$

Time for first half of the distance

For the first half, distance covered is $s=\frac{h}{2}$

Hence, $\frac{h}{2}=\frac{1}{2}gt_1^2$

$h=gt_1^2$

$t_1=\sqrt{\frac{h}{g}}$

Total time to fall the full height $h$

Let total time be $T$. Then $h=\frac{1}{2}gT^2$

$T=\sqrt{\frac{2h}{g}}$

Time for second half of the distance

The second half time is $t_2=T-t_1$

So, $t_2=\sqrt{\frac{2h}{g}}-\sqrt{\frac{h}{g}}$

$t_2=\sqrt{\frac{h}{g}}(\sqrt{2}-1)$

But from step 2, $t_1=\sqrt{\frac{h}{g}}$

Therefore, $t_2=(\sqrt{2}-1)t_1$

Compare with options

This matches: $\boxed{t_2=(\sqrt{2}-1)t_1}$

So the correct option is D.