Standard Method
When particle #1 reaches the point of collision, its downward speed $u_1$ is given by $$u_1^2=2g(\frac12 h)=gh$$
Particle #2 covers the same distance in the same time, so it has the same average velocity $\frac12 u_1$. Its change in velocity (a reduction) is also the same, ie $u_1$. So its initial velocity must have been $u_1$ and its final velocity (at collision) is $u_2=0$.
The downward speed $v$ of the combined mass after the completely inelastic collision is found from the conservation of momentum :$$3mv=mu_1+3mu_2$$ $$v=\frac13 u_1$$
The combined particle reaches the ground with speed $w$ given by $$w^2=v^2+2g(\frac12h)=\frac19 u_1^2+gh=\frac{10}{9}gh$$ $$w=\frac13\sqrt{10gh}$$
Centre of Mass Method
The CM follows the same trajectory whether the particles collide inelastically, elastically or not at all. The loss of KE during the collision is an "internal" change which does not affect the motion of the CM.
This method looks like it might be a smart way to get the answer quickly. It would be quicker if we knew the initial velocity $u$ of the CM, but we don't - we have to find this.
The time taken for the two particles to collide is the time for particle #1 to fall from rest a distance of $\frac12 h$. So this time $t$ is given by $\frac12 h=\frac12 gt^2$ so $t^2=\frac{h}{g}$.
If particle #2 has initial velocity $u_2$ upwards then $h=u_2 t$ so $u_2^2=gh$. Particle #1 has initial velocity $u_1=0$ so the initial velocity of the CM is given by $$u=\frac13u_1+\frac23u_2$$ $$u^2=\frac49u_2^2=\frac49gh$$
The CM is initially at height $\frac13 h$ above ground so its velocity $w$ when it reaches the ground is given by $$w^2=u^2+2g(\frac13 h)=\frac49gh+\frac23 gh=\frac{10}{9}gh$$ $$w=\frac13 \sqrt{10gh}$$
This has been no easier than the Standard Method.
