Two balls falling one above the other [closed]

Two balls, first with the mass $m_1$ and the second with the mass $m_2$ are falling from the heigh $h$. Suppose all the collisions are perfectly elastic and do not consider the size of the balls. $m_1 < m_2$ and ball with the mass $m_1$ is on the top.

What height $h_1$ will the ball with the mass $m_1$ jump to and what height $h_2$ will the ball with the mass $m_1$ jump to?

I think equations as $(m_1+m_2)gh=m_1gh_1+m_2gh_2$ and $(m_1+m_2)v=m_1v_1+m_2v_2$, where $v$ is the velocity of both balls falling, are correct, but I have no idea how to get $h_1$ and $h_2$ independent of each other.

It seems to me, like here is some common operation or relation, that I just don't know, that I have never seen or don't remember... Please don't describe me "you have to do this and this" and "think about what happen when...". I need to see the progress. I need to see these equations and what's happening.

-

closed as off-topic by akhmeteli, John Rennie, user1504, twistor59, Manishearth♦Jun 26 '13 at 22:55

This question appears to be off-topic. The users who voted to close gave this specific reason:

If this question can be reworded to fit the rules in the help center, please edit the question.

First, figure out how much energy is lost by the two balls as they fall to the ground. Now, the first ball reverses its momentum upon hitting the ground. Now, you have the one ball going toward the ground with speed $v$ and the other ball going upward at speed $v$.

What happens when two balls collide elastically with a head-on collision of speed $v$ in opposite directions? After the collision, how much energy does ball one have? How much energy does ball two have? What does this tell you about their final heights? How can you write these heights in terms of the initial height?

-
Did you try to compute it? Did you get the results i wrote it should have? Cause I tried everything all these people said and nothing led me to the right answer. –  user50222 Aug 27 '13 at 3:22
@user50222: yes. It worked out to the result you have in your comment to Sahil Chada's answer. –  Jerry Schirmer Aug 27 '13 at 14:01
This is not the right answer. Result is $h_1=\left( \frac{3m_2-m_1}{m_1+m_2} \right)^2h$ and $h_2=\left( \frac{m_2-3m_1}{m_1+m_2} \right)^2h$ but I don't know how to get these. –  user50222 Jun 20 '13 at 7:36
@user50222: what happens to those two formulae when $m_{1} = m_{2}$? –  Jerry Schirmer Jun 20 '13 at 8:59