Answer
$a_{com} = (-3.675~m/s^2)~\hat{j}$
Work Step by Step
Since Particle 2 always stays directly above Particle 1, they must have the same constant horizontal velocity of $10.0~m/s$.
Throughout the motion, the horizontal component of acceleration of both particles is $0$
The vertical component of acceleration of Particle 1 is $0$ while the vertical component of acceleration of Particle 2 is $-9.8~m/s^2$
We can find the vertical component of acceleration of the center of mass:
$a_{com,y} = \frac{(5.00~g)(0)+(3.00~g)(-9.8~m/s^2)}{5.00~g+3.00~g}$
$a_{com,y} = -3.675~m/s^2$
We can write the acceleration of the center of mass in unit-vector notation:
$a_{com} = (-3.675~m/s^2)~\hat{j}$
