# Centre Of Gravity Help!

1. Nov 19, 2007

### cfc101

1. The problem statement, all variables and given/known data

A small ball of mass m is dropped at a height of 50 metres. At the same time, a second ball with twice the mass of the first one is launched at an initial velocity of 10m/s from the ground. Where is the centre of gravity in this system?

2. Relevant equations

3. The attempt at a solution
I dont really know where to begin on this one, although i know that the cener of mass will probably lie closer to the ground because of heavier ball. All help is appreciated, thanks in advance.

2. Nov 19, 2007

### Staff: Mentor

How does one define center of mass, given the position of two masses in the y-direction?

What is happening with the position of those masses?

See if this helps - http://hyperphysics.phy-astr.gsu.edu/hbase/traj.html

Remember if something is falling and it's position is y(t) from some initial height h from the ground, then it's height or elevation is given by h-y(t).

Last edited: Nov 19, 2007
3. Nov 19, 2007

### cfc101

Well, since this problem only deals with the y axis, the center of mass is 1/3h off the ground (if stationary, using the mass ratio), however, I dont really understand how velocity is factored into all of this

4. Nov 19, 2007

### Staff: Mentor

The positions (elevations) of each mass are changing with time.

ycm(t) = (y1(t)m1+y2(t)m2)/(m1+m2), so determine yi(t).

What are the positions y1(t) and y2(t) as functions of inital position, velocity, acceleration and time.

Last edited: Nov 19, 2007
5. Nov 20, 2007

### cfc101

h = h0 + Vot + 1/2 t^2
so for the first mass at the top

h1 = 50 -4.9t^2
mass at bottom

h2 = 10t -4.9t^2

To get the center of mass, would i need to add these two equations?

Cm = -9.8t^2 + 10t + 50 -----> -4.9t + 5t + 25

I have no idea what to do from here

6. Nov 20, 2007