Strategy for a person throwing an object in zero G to maximize a velocity If an astronaut is drifting away from a space station holding a wrench, they could throw it in the opposite direction in an attempt to get back.  But if they do it wrong, they will only end up spinning in place.  Also, humans are much better at throwing things in certain directions than in others.
Knowing that clever maneuvers allow someone to change their orientation in space, as long as they don't end up with a net angular momentum or net momentum, what is the best strategy for a baseball pitcher to achieve the highest speed possible in a chosen direction?
 A: An ideal situation would be that the wrench's final direction of travel is in a straight line through the astronaut's center of gravity and his desired destination, but ones' body is in the way.  Overhand throws are problematic.  One should try to push the ball away, or fling it with a back-hard motion.  If throwing directly up or directly down, one should keep the wrench very close to the body, and "fishtail" their throw once past, in order to line these up.
A: Throwing a wrench is the same problem as being impacted by a wrench in space. Most efficient transfer of momentum is via the center of mass. You can write the body momentum equation as 
$$\boldsymbol{p} = \boldsymbol{n}\, J \tag{1}$$ 
Where $\boldsymbol{n}$ is the direction of travel and $J$ the impulse needed.
This applies as equal and opposite sense on astronaut and wrench. The required impulse is calculated from 
$$ J = m_{\rm eff}\, u_{\rm throw} \tag{2} $$ 
Where $u_{\rm throw}$ is the known relative speed one can throw a wrench, and $m_{\rm eff}$ the effective mass of the system. This, in turn, is calculated from each mass $m_{\rm A}$ and $m_{\rm W}$ as well as the mass moment of inertia (tensor) of the astronaut $\mathbf{I}_{\rm a}$ and the point of release relative to the center of mass $\boldsymbol{r}$.
$$ \frac{1}{m_{\rm eff}} = \frac{1}{m_{W}}+\frac{1}{m_{A}} - \boldsymbol{n}^\top \left(\boldsymbol{r}\times\mathbf{I}_{A}^{-1}\left(\boldsymbol{r}\times\boldsymbol{n}\right)\right) \tag{3}  $$
The final (linear and angular) motion of the astronaut is given by the following expressions
$$ \begin{aligned}
\Delta\boldsymbol{v}_{A}&=\frac{\boldsymbol{n}J}{m_{A}}=\boldsymbol{n}\left(\frac{m_{{\rm eff}}}{m_{A}}\right)u_{{\rm throw}}\\\Delta\boldsymbol{\omega}_{A}&=\mathbf{I}_{A}^{-1}\left(\boldsymbol{r}\times\boldsymbol{n}\right)J=\mathbf{I}_{A}^{-1}\left(\boldsymbol{r}\times\boldsymbol{n}\right)m_{{\rm eff}}u_{{\rm throw}}
\end{aligned} \tag{4}$$
Regardless of the rotation, the most efficient throw is one that maximizes the ratio $$\frac{m_{{\rm eff}}}{m_{A}}=\frac{1}{\frac{m_{A}}{m_{W}}-\boldsymbol{n}^\top \left(\boldsymbol{r}\times\mathbf{I}_{A}^{-1}\left(\boldsymbol{r}\times\boldsymbol{n}\right)\right)m_{A}} \tag{5}$$
This is done by maximizing the wrench mass $m_W$ and making the throw point on the center of mass $\boldsymbol{r}=0$. This also has the effect of zero rotation for the astronaut making his approach more controllable.
So in summary, take the heaviest thing you can throw, and translate it away from your center of mass in the direction opposite of where you want to go.
