Using conservation of energy we find the velocity of ball

A

just before impact when it has fallen a vertical distance

4 h

2 mg (5 h - h) = \frac{1}{2} . 2 m . u_{A}^{2}

u_{A} = \sqrt{8} gh .

By conservation of momentum at the impact, taking the velocities of the balls before and after impact to be

u_{A}, v_{A}

and

u_{B} = 0, v_{B}

2 {mu}_{A} = 2 {mv}_{A} + {mv}_{B}

and using the coefficient of restitution:

0.5 u_{A} = v_{B} - v_{A}

From these two equations:

v_{B} = u_{A} = \sqrt{8} gh

and

v_{A} = 0.5 u_{A}

The time $t$ before $B$ hits the ground is given by:

$h = \frac{1}{2} {gt}^{2}$

so

$t = \sqrt{\frac{2 h}{g}} .$

The horizontal distance traveled in this time is given by:

$d = \sqrt{8} gh \times \sqrt{\frac{2 h}{g}} = 4 h .$