Let the speeds of the two balls ( 1 and 2 ) be $u_1$ and $u_2$ where if $u_1=2 u_2$ and $u_1=u_2=u$.
If $h_1$ and $h_2$ and the distance covered by the balls 1 and 2 , respectively, before coming to rest, then
$h_1=\frac{u_1^2}{2 g}=\frac{4 u^2}{2 g}$ and $h_2=\frac{u_2^2}{2 g}=\frac{u^2}{2 g}$
According to question,
$$
\begin{aligned}
&\left(h_1-h_2\right)=15 \mathrm{~m} \\
&\frac{4 u^2}{2 g}-\frac{u^2}{2 g}=15 \mathrm{~m} \text { or } \frac{3 u^2}{2 g}=15 \mathrm{~m} \\
&\text { or } u^2=\sqrt{5 \mathrm{~m} \times(2 \times 10)} \mathrm{m} / \mathrm{s}^2 \\
&u=10 \mathrm{~m} / \mathrm{s} \\
&
\end{aligned}
$$
So velocity of ball 1 and $2, u_1=20 \mathrm{~m} / \mathrm{s}$ and $u_2=10 \mathrm{~m} / \mathrm{s}$
As, $h_1=\frac{u_1^2}{2 g}=\frac{(20 \mathrm{~m})^2}{2 \times 10 \mathrm{~m} 15}=20 \mathrm{~m}$
$$
h_2=h_1-15 \mathrm{~m}=5 \mathrm{~m}
$$
For 1st ball, $v_1=u_t+g t_1$
$$
\begin{aligned}
&0=20-10 t_1 \\
&t_1=2 \mathrm{sec} .
\end{aligned}
$$
For second ball, $v_2=u_2+g t_2$
$$
\begin{aligned}
&0=10-10 t_2 \\
&t_2=1
\end{aligned}
$$
Exact time interval between the two balls
$$
=t_1-t_2=2 \mathrm{~s}-1 \mathrm{~s}=1 \text { second. }
$$