Answer:
the center of mass is 316,670m bellow the release point.
Explanation:
First, we must find the distance at which the objects are in time t = 360s
We will use the formula for vertical distance in free fall
![h=v_{0}t+\frac{1}{2} gt^2](https://tex.z-dn.net/?f=h%3Dv_%7B0%7Dt%2B%5Cfrac%7B1%7D%7B2%7D%20gt%5E2)
is the initial velocity, is 0 for both stones since they were just dropped.
g is acceleration of gravity and t is time (
)
At
the first stone has been falling for the entire 360 seconds, its position h1 is:
![h_{1}=\frac{1}{2} (9.82m/s^2)(360s^2)=635,688m](https://tex.z-dn.net/?f=h_%7B1%7D%3D%5Cfrac%7B1%7D%7B2%7D%20%289.82m%2Fs%5E2%29%28360s%5E2%29%3D635%2C688m)
And at 360 seconds the second stone has been fallin for
, so its position h2 is:
![h_{2}=\frac{1}{2} (9.81m/s^2)(179)^2=157,161.1m](https://tex.z-dn.net/?f=h_%7B2%7D%3D%5Cfrac%7B1%7D%7B2%7D%20%289.81m%2Fs%5E2%29%28179%29%5E2%3D157%2C161.1m)
And finally using the equation for the center of mass:
![CM=\frac{m_{1}h_{1}+m_{2}h_{2}}{m_{1}+m_{2}}](https://tex.z-dn.net/?f=CM%3D%5Cfrac%7Bm_%7B1%7Dh_%7B1%7D%2Bm_%7B2%7Dh_%7B2%7D%7D%7Bm_%7B1%7D%2Bm_%7B2%7D%7D)
We know that the mass of the second stone is twice the mass of the first stone so:
![m_{1}=m\\m_{2}=2m](https://tex.z-dn.net/?f=m_%7B1%7D%3Dm%5C%5Cm_%7B2%7D%3D2m)
replacing these values in the equation for the center of mass
![CM=\frac{mh_{1}+2mh_{2}}{m+2m}](https://tex.z-dn.net/?f=CM%3D%5Cfrac%7Bmh_%7B1%7D%2B2mh_%7B2%7D%7D%7Bm%2B2m%7D)
![CM=\frac{m(h_{1}+2h_{2})}{3m}=\frac{h_{1}+2h_{2}}{3}](https://tex.z-dn.net/?f=CM%3D%5Cfrac%7Bm%28h_%7B1%7D%2B2h_%7B2%7D%29%7D%7B3m%7D%3D%5Cfrac%7Bh_%7B1%7D%2B2h_%7B2%7D%7D%7B3%7D)
Finally, replacing the values we found fot h1 and h2:
![CM=\frac{635,688m+2(157,161.1m)}{3}=316,670m](https://tex.z-dn.net/?f=CM%3D%5Cfrac%7B635%2C688m%2B2%28157%2C161.1m%29%7D%7B3%7D%3D316%2C670m)
the center of mass is 316,670m bellow the release point.