It takes t 1 s to set the tower height to x A and m meters for free fall, t2 seconds for total * * *, and t2-t 1 for b fall.

m=gt 1？ /2

x=gt2？ /2

x-n=g(t2-t 1)？ /2

Solve it

T2 = (m+n)/√ (2mg)

x=(m+n)？ /(4 meters)

The limit case is that when the speed of ball B drops to 2m/s, the two balls do not collide.

Because when the speed of B is less than 2m/s, it is impossible for two balls to hit.

Let the speed of b be v.

v-2.5t=2

B The average speed during this period is (v+2)/2.

(v+2)/2 * t - 2t =20

t=4

v= 12