Question: I'm trying to figure out which object would have the largest time of flight when launched from a slingshot. To do this, I need the initial velocity. How can I calculate this if I know the mass and diameter of the object? The slingshot always has the same force. I've seen a number of formulas, but none of them take into account the mass and size, which would matter in this case (a 10kg, 10cm rock will not go as fast or as long as a 1kg, 1cm rock).

Answer: 
Your problem is not related only with the Dirac equation, but with whatever field equation you have. Your question is about the interpretation of the field as a wave function of one particle. There are two concepts that I would like to show here that might answer your question: fields and wave functions (they are different things).

* A field is an object $\phi\_{\alpha} (x)$, where $x$ behaves as a event in space time and $\alpha$ is an index that carry a [finite representation of Lorentz Symmetry](https://en.m.wikipedia.org/wiki/Representation_theory_of_the_Lorentz_group#Finite-dimensional_representations). They obey some equations that related the $\alpha$ indices and the $x$, but this is irrelevant here. What is relevant is that they also obey the Klein Gordon equation, which is given by
$$
\partial^2\phi\_{\alpha}= \pm m^2\phi\_{\alpha},
$$
for the metric signature ($\pm,\mp, \cdots,\mp $).
* A wave function is a representation of a quantum state in a particular bases.
$$
\psi (x)=\langle x|\psi\rangle
$$
where $|x\rangle$ behave as a event (in non-relativistic quantum mechanics this is $(t,\vec {x}) $). Turns out that this state $|x\rangle$ in relativistic quantum mechanics is:
$$
|x\rangle=\int\frac{d^3p}{(2\pi)^{3/2}}\sqrt{\frac{k^0}{p^0}}e^{-ipx}|p\rangle
$$
You can prove yourself as an exercise. Note that there is a square root, which prevents this integral to be a simple Fourier transform of $|p\rangle$, the eigenvector of P (generator of translations).
The set of states $|x\rangle$ does not constitute an orthogonal basis since
$$
\langle x|y \rangle \neq 0
$$ 
and the square root in the formula above is the responsible for that. So, if we work with a wave function $\langle x|\psi\rangle$ in a relativistic QM we are dealing with am over-complete basis representation.

Now if we upgrade our language and move to the [second-quantization](https://en.wikipedia.org/wiki/Second_quantization) framework we have 
$$
|x\rangle=a^{\dagger}\_x|0\rangle
$$
Defining linear combinations of creation and annihilation operators of the form:
$$
\phi(x)=Aa\_x+Ba\_x^{\dagger}
 $$
such that $[\phi(x),\phi^{\dagger}(y)]= 0$ at space like separations is a useful construction to model local interactions of relativistic QM. This leads to the [Quantum Field Theory](https://en.wikipedia.org/wiki/Quantum_field_theory) enterprise.
As showed by Dirac and others QFT is a *disjoint* theory of RQM; as a consequence, QFT cannot solve the problems of RQM. You must want to revise the section "8.3 Does QFT solve the problems of relativistic QM?" of the *Phys. Found.* paper [Quantum mechanics: Myths and facts](http://arxiv.org/abs/quant-ph/0609163). This is the conclusion:

> 
> Thus, instead of saying that QFT solves the problems of relativistic QM, it is more honest to say that it merely sweeps them under the carpet.
> 
> 
>