A, B, A do vertical throwing, and if the vertical upward direction is taken as the positive direction, there are: -h=v0ta- 12gt2a,

Solution: ta=v0+v20+2ghg.

When B throws vertically downward, there are: 2h=v0tb+ 12gt2b,

Solution: tb=v20+4gh? v0g

If ta=tb, the simultaneous solution is V 0 = GH 12. When V 0 > GH 12, the falling time of ball b is shortened, while the falling time of ball a is prolonged, so ball b lands first. On the other hand, when V 0 < GH 12, the ball A can land first, so AB is wrong.

C. Before landing, ball A and ball B are only subjected to gravity, and mechanical energy is conserved, so C is correct.

D. When both balls are moving, the kinetic energy is equal. If the gravitational potential of ball B is large, the mechanical energy of ball B is large, and the mechanical energy of both balls is conserved during the air movement, so the mechanical energy of ball A is smaller than that of ball B before and after the encounter, so D is wrong.

So choose: C.