#### Answer

Ed rises to $1.54m$ above his starting point.

#### Work Step by Step

Before they push each other, Adolf and Ed were hanging at rest so $\sum p_0=0$
After pushing, Adolf, with mass $M=120kg$, has velocity $V$, while Ed, with mass $m=78kg$, has velocity $v$. We take Adolf's direction of velocity to be positive, so Ed's velocity is negative.
According to the principle of conservation of linear momentum, $$MV-mv=\sum p_0=0$$ $$MV=mv$$ $$\frac{V}{v}=0.65 (1)$$
1) Now we look at Adolf's swinging upward:
After pushing, he swings upward to a height $H=0.65m$, at which points his velocity $V_f=0$. Following the conservation of mechanical energy, we have $$\frac{1}{2}M(V_f^2-V^2)+MgH=0$$ $$\frac{1}{2}(V_f^2-V^2)+gH=0$$ $$\frac{1}{2}(0-V^2)+9.8\times0.65=0$$ $$\frac{1}{2}V^2=6.37$$ $$V=3.57m/s$$
Plugging into (1), we can find Ed's velocity right after they push each other: $v=V/0.65=5.49m/s$
2) Now we look at Ed's swinging upward:
After pushing, Ed has velocity $v=5.49m/s$ and swings upward to a height $s$, at which points his velocity $v_f=0$. Following the conservation of mechanical energy, we have $$\frac{1}{2}m(v_f^2-v^2)+mgh=0$$ $$\frac{1}{2}(v_f^2-v^2)+gh=0$$ $$h=\frac{v^2-v_f^2}{2g}=1.54m$$