Answer
The center of mass of the two stones is a distance of $~~0.28~m~~$ below the release point.
Work Step by Step
We can find the distance the first stone falls in $0.30~s$:
$y = \frac{1}{2}at^2$
$y = \frac{1}{2}(9.8~m/s^2)(0.30~s)^2$
$y = 0.441~m$
We can find the distance the second stone falls in $0.20~s$:
$y = \frac{1}{2}at^2$
$y = \frac{1}{2}(9.8~m/s^2)(0.20~s)^2$
$y = 0.196~m$
Let $M$ be the mass of the first stone.
Let $2M$ be the mass of the second stone.
We can find the location of the center of mass below the release point:
$y = \frac{(M)(0.441~m)+(2M)(0.196~m)}{M+2M}$
$y = \frac{0.833~M}{3~M}$
$y = 0.28~m$
The center of mass of the two stones is a distance of $~~0.28~m~~$ below the release point.
