From the meaning of the question, it can be concluded that when M touches the seesaw, the moment of the system to C is zero, so the angular momentum is conserved.

Initial angular momentum (only m has momentum from m to the seesaw) mv 1L/2 v 1? =2gh (free fall)

The angular momentum of the final momentum (when n leaves) n is mv2L/2.

The angular momentum of m is mv2L/2, and the inelastic velocity of two people is always the same.

The angular momentum of the seesaw needs to be integrated to find 2∫(2mv2/L? )r？ Dr (lower limit 0, upper limit L/2)

=mv2L/6

Initial angular momentum = final momentum

mv 1L/2 = mv2L/2+mv2L/2+mv2L/6v 1？ =2gh

v2=3v 1/7=3√(2gh)/7

Height of n h 1 2gh 1=v2? = 18gh/49

h 1=9h/49