Answer
$H_{max} = 5.74~m$
Work Step by Step
Since Particle 2 always stays directly above Particle 1, they must have the same horizontal velocity.
We can find the angle $\theta$ above the horizontal at which Particle 2 is launched:
$v_x = (20.0~m/s)~cos~\theta = 10.0~m/s$
$cos~\theta = \frac{10.0~m/s}{20.0~m/s}$
$\theta = cos^{-1}~(\frac{1}{2})$
$\theta = 60^{\circ}$
We can find the vertical component of the initial velocity of Particle 2:
$v_{0y} = (20.0~m/s)~sin~60^{\circ}$
$v_{0y} = 17.32~m/s$
We can find the maximum height of Particle 2:
$v_{fy}^2 = v_{0y}^2+2ay$
$y = \frac{v_{fy}^2 - v_{0y}^2}{2a}$
$y = \frac{0 - (17.32~m/s)^2}{(2)(-9.8~m/s^2)}$
$y = 15.3~m$
We can find the maximum height of the center of mass:
$H_{max} = \frac{(5.00~g)(0)+(3.00~g)(15.3~m)}{5.00~g+3.00~g}$
$H_{max} = 5.74~m$
