Let us assume the mass of the ball is $3m$ and the velocity of the lighter piece in horizontal direction after the explosion is $v$. We know that just after the collision, the horizontal velocity of the heavier piece is $10 \mathrm{m} {\mathrm{s}}^{-1}$. Using conservation of momentum in the horizontal direction we get

$mv=2m\times 10$

$\Rightarrow v=20 \mathrm{m} {\mathrm{s}}^{-1}$

Since there is no change in vertical motion, the time taken by the pieces to reach the ground can be calculated in the following way.

Total time to fall through $45 \mathrm{m}$ is

$T_{45}=\sqrt{\frac{2\times 45}{10}}=3 \mathrm{s}$

The time taken to fall through first $20 \mathrm{m}$ is

$T_{20}=\sqrt{\frac{2\times 20}{10}}=2 \mathrm{s}$.

Hence time taken by the pieces to fall from $25 \mathrm{m}$ height to ground is

$T_{40}-T_{20}=3-2=1 \mathrm{s}$

The relative velocity of the pieces in the horizontal direction is

$v_{\mathrm{rel}}=v-\left(-10\right)=30 \mathrm{m} {\mathrm{s}}^{-1}$

The horizontal distance between the two pieces at the time of striking the ground is

$x=30\times 1=30 \mathrm{m}$