A ball starts falling under the effect of gravitational force from...

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 $10ms^{- 1}$. Using conservation of momentum in the horizontal direction we get

$mv = 2m \times 10$

$⇒ v = 20ms^{- 1}$

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 $45m$ is

${T}_{45} = \sqrt{\dfrac{2 \times 45}{10}} = 3s$

The time taken to fall through first $20m$ is

${T}_{20} = \sqrt{\dfrac{2 \times 20}{10}} = 2s$.

Hence time taken by the pieces to fall from $25m$ height to ground is

${T}_{40} - {T}_{20} = 3 - 2 = 1s$

The relative velocity of the pieces in the horizontal direction is

${v}_{rel} = v - \left(- 10\right) = 30ms^{- 1}$

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

$x = 30 \times 1 = 30m$