Let the speeds of the two balls $v_{1}$ and $v_{2}$ respectively.

If $v_{2}=v$, then $v_{1}=2 v$

If $y_{1}$ and $y_{2}$ are the distance covered by the balls respectively before coming to rest, then

$y_{1}=\frac{v_{1}^{2}}{2 g}=\frac{4 v^{2}}{2 g} \text { and } y_{2}=\frac{v_{2}^{2}}{2 g}=\frac{v^{2}}{2 g}$

$y_{1}-y_{2}=15 \mathrm{~m}$

$\frac{4 v^{2}}{2 g}-\frac{v^{2}}{2 g}=15 \mathrm{~m}$

$\frac{3 v^{2}}{2 g}=15 \mathrm{~m}$

$\therefore v^{2}=\sqrt{5 \mathrm{~m} \times(2 \times 10)} \mathrm{m} / \mathrm{s}^{2}$

$\therefore v=10 \mathrm{~m} / \mathrm{s}$

$\therefore v_{2}=10 \mathrm{~m} / \mathrm{s}$ and $v_{1}=20 \mathrm{~m} / \mathrm{s}$

Now, $y_{1}=\frac{v_{1}^{2}}{2 g}=\frac{(20 \mathrm{~m})^{2}}{2 \times 10 \mathrm{~m} 15}=20 \mathrm{~m}$

$y_{2}=y_{1}-15 \mathrm{~m}=5 \mathrm{~m}$

If $t_{2}$ is the time taken by the ball 2 to cover a distance of $5 \mathrm{~m}$, then,

From $y_{2}=v_{2} t-\frac{1}{2} g t_{2}^{2}$

$ 5=10 t_{2}-5 t_{2}^{2} \text { or } t_{2}^{2}-2 t_{2}+1=0$

$\therefore  t_{2}=1 \mathrm{~s}$

Now, $t_{1}$ (time taken by ball 1 to cover distance of $20 \mathrm{~m}$ ) is $2 \mathrm{~s}$, time interval between the two throws $=t_{1}-t_{2}=2 \mathrm{~s}-1 \mathrm{~s}=1 \mathrm{~s}$