Given: Height of free fall of ball is  $H$

Distance covered by the ball is $\frac{H}{2}$ in last $1.0 \mathrm{s}$.

Now, displacement in the $n^{th}$ second $s_n=u+a\left(n-\frac{1}{2}\right)$

Here $s_n=\frac{H}{2}$, $u=0$ (since the ball is at rest initially)

Thus, $\frac{H}{2}=g\left(n-\frac{1}{2}\right) ...\left(1\right)$

where $n$ is the time taken to reach the ground $n=\sqrt{\frac{2H}{g}}$

$\frac{n^{2}g}{2}=H$

Substituting the value of $n$ in equation $\left(1\right)$

$\frac{n^{2}g}{2\times 2}=g\left(n-\frac{1}{2}\right)$

$n^2=4n-2$

$n^2-4n+2=0$

$n=2\pm \sqrt{2}$

$n$ should be greater than $1$

so, $n=2+\sqrt{2}=3.41 \mathrm{s}$