Using conservation of energy we find the velocity of ball A just before impact when it has fallen a vertical distance 4h:

2mg(5h-h) =

.2m. u_A²

so u_A = Ö8gh. 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:

2mu_A = 2mv_A + mv_B

and using the coefficient of restitution:

0.5u_A = v_B - v_A

From these two equations: v_B = u_A = Ö8gh and v_A = 0.5 u_A.

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

h = 1
2
gt²

so

t = æ
ú
Ö

2h
g

.

The horizontal distance traveled in this time is given by:

d = Ö8gh × æ
ú
Ö

2h
g

= 4h.