Answer
a. 7d/3.
b. 5d.
Work Step by Step
The missile’s center of mass follows the same path no matter how the mass splits, and lands at a distance of 2d from the starting point.
Assume that $m_I$ is stopped in midair and falls straight down, as in the given example.
$$x_{CM}=\frac{(m_I)(x_I)+ (m_{II})(x_{II}) }{m_I+m_{II}}$$
a.
$$x_{CM}=2d=\frac{(m_I)(d)+ (3m_{I})(x_{II}) }{3m_I}$$
$$2d=\frac{d+ 3(x_{II}) }{3}$$
$$7d=3x_{II}$$
$$x_{II}=\frac{7}{3}d$$
b.
$$x_{CM}=2d=\frac{(3m_{II})(d)+ (m_{II})(x_{II}) }{4m_{II}}$$
$$2d=\frac{3d+ (x_{II}) }{4}$$
$$x_{II}=5d$$
