I am going to change the notation in order to make the equations more compact. The counter weight is $M$ and the payload is $m$. The length of the bar is $L$ and the distance of the center of gravity to the counterweight is $a=\frac{m}{M+m}\,L$ and from the payload $b=\frac{M}{M+m}\,L$ such that $L=a+b$. Note I have said nothing about the pivot yet.
The distance between the pivot and the center of gravity is $c$ and it is an independent variable we wish to optimize. The pivot is between the center of gravity and the payload (for positive $c$). The angle of the bar is $\theta$ with $\theta=0$ when horizontal.
The height of the pivot from the ground is $h$ such that when the counterweight hits the ground the payload is launched at $\theta_{f} = -45^\circ$. So $h=(a+c) \sin(- \theta_f$). As a consequence the initial angle is $\sin \theta_i = \frac{a+c}{b-c} \sin(- \theta_f )$ in order for the payload to rest in the ground initially. This is valid for $c<\frac{b-a}{2}$, otherwise the things sits vertically with $\theta_i=\frac{\pi}{2}$.
Doing the dynamics using Newtons's Laws, or Langrange's equations will yield the following acceleration formula
$$ \ddot{\theta} = -\,\frac{c g (M+m) \cos(\theta)}{\frac{m M}{M+m} L^2 + (M+m) c^2} $$
The denominator being the moment of inertia about the pivot. Here is the fun stuff.
The above can be integrated since the right hand side is a function of $\theta$ only with a constant $\alpha$:
$$ \ddot{\theta}=\frac{{\rm d}\dot\theta}{{\rm d}t} =-\alpha \cos(\theta) $$
$$ \frac{{\rm d}\dot\theta}{{\rm d}\theta} \frac{{\rm d}\theta}{{\rm d}t} = -\alpha \cos(\theta) $$
$$ \frac{{\rm d}\dot\theta}{{\rm d}\theta} \dot\theta = -\alpha \cos(\theta) $$
$$ \int \dot\theta {\rm d}\dot\theta =-\int \alpha \cos(\theta) {\rm d}\theta + K$$
$$ \frac{{\dot \theta}^2}{2} = -\alpha \sin \theta + K $$
with $K$ based on the initial conditions ($\theta=\theta_i$, $\dot\theta=0$)
$$ \dot\theta = \sqrt{2 \alpha (\sin(\theta_i)-\sin(\theta))}$$
and final rotational velocity
$$ \dot\theta_f = \sqrt{2 \alpha (\sin(\theta_i)-\sin(\theta_f))}$$
tangentially the payload launch velocity is
$$ v_{B_f} = (b-c) \dot\theta_f = (b-c) \sqrt{2 \alpha (\sin(\theta_i)-\sin(\theta_f))} $$
with both $\alpha$ and $\theta_i$ depending on the variable $c$.
To optimize we set $ \frac{{\rm d}v_{B_f}}{{\rm d}c}=0 $ which is solved for:
$$ \frac{c}{L} = \frac{\sqrt{m} \left( \sqrt{M+m}-\sqrt{m} \right)}{M+m } $$
For example, a $m=20 {\rm lbs}$ payload, with a $M=400 {\rm lbs}$ counter weight on a $L=20 {\rm ft}$ bar, requires the pivot to be $c=20\;\frac{\sqrt{20} \left( \sqrt{420}-\sqrt{20} \right)}{420 } = 3.412\;{\rm ft} $ from the center of gravity. The c.g. is $a=\frac{20}{420}\,20 = 0.952 {\rm ft}$ from the counterweight.
Edit 1
Based on comments made by the OP the launch velocity is
$$
v_{B_{f}}=\left(\frac{M}{M+m}L-c\right)\sqrt{\frac{2cg(M+m)}{\frac{M}{M+m}L^{2}+(M+m)c^{2}}\left(\sin\theta_{i}-\sin\theta_{f}\right)}
$$
where $g=9.80665\,{\rm m/s}$ is gravity.
With infinite counterweight the maximum launch velocity is $\max(v_{B_{f}})=\sqrt{\frac{2g(L-c)^{2}}{c}}$ so to reach $v_{B_{f}}=6000\,{\rm m/s}$ from earth if $c=1\,{\rm m}$ then $L>1355.8\,{\rm m}$.
With infinite bar length the maximum launch velocity is $\max(v_{B_{f}})=\sqrt{\frac{2Mcg}{m}}$ so to reach $v_{B_{f}}=6000\,{\rm m/s}$ from earth if $c=1\,{\rm m}$ then $M>18.3\,10^{6}\,{\rm kg}$.
So lets consider $L=2000\,{\rm m}$ and $M=40.0\,10^{6}\,{\rm kg}$ then we choose pivot location at $c=1.500\,{\rm m}$ to get
$$
v_{B_{f}}=(1999.999-1.5)\sqrt{2\;4.526\left(\sin\theta_{i}-\sin\theta_{f}\right)}
$$
which is solved for $v_{B_{f}}=6000\,{\rm m/s}$ when $\sin\theta_{i}-\sin\theta_{f}=0.9957$
with $\theta_{i}>0$ and $\theta_{f}<0$.
Best Answer
In general air resistance makes the optimal angle lower than 45$^{\circ}$. However I don't know of any rigorous way to show this without sitting down at the computer. The argument usually used is that in the prescence of air resistance you want to minimise the energy lost to the air by reducing the time of flight, and this means choosing a lower trajectory. The reduction in range by using a lower angle is balanced out by the greater average speed of the projectile.
If you Google for "optimum projectile angle air resistance" or something like this you'll find lots of articles going into this question.