Estimation of the bond angle of water
how would people give estimates of the actual bonding angle of water
What physics would be involved in the calculation

Background
That's a very good question. In many cases Coulson's Theorem can be used to relate bond angles to the hybridization indices of the bonds involved.
$$\ce{1+\lambda_{i} \lambda_{j} cos(\theta_{ij})=0}$$
where $\ce{\lambda_{i}}$ represents the hybridization index of the $\ce{C-i}$ bond (the hybridization index is the square root of the bond hybridization) and $\ce{\theta_{ij}}$ represents the $\ce{i-C-j}$ bond angle.

For example, in the case of methane, each $\ce{C-H}$ bond is $\ce{sp^3}$ hybridized and the hybridization index of each $\ce{C-H}$ bond is $\sqrt3$. Using Coulson's theorem we find that the $\ce{H-C-H}$ bond angle is 109.5 degrees.
How can we use this approach to answer your question?
Unlike methane, water is not a perfectly tetrahedral molecule, so the oxygen bonding orbitals and the oxygen lone pair orbitals will not be exactly $\ce{sp^3}$ hybridized.  Since addition of s-character to an orbital stabilizes electrons in that orbital (because the s orbital is lower in energy than the p orbital) and since the electron density is higher in the lone pair orbital than the $\ce{O-H}$ orbital (because the electrons in the lone pair orbital are not being shared with another atom), we might expect that oxygen lone pair orbital will have more s-character and the oxygen $\ce{O-H}$ orbital will have less s-character.
If we examine the case where the $\ce{O-H}$ bond is $\ce{sp^4}$ hybridized, we find from Coulson's Theorem that the $\ce{H-O-H}$ angle is predicted to be around 104.5°. So indeed, in agreement with our prediction, removing s-character from the oxygen $\ce{O-H}$ orbital gives rise to the observed bond angle.
Note on reality
For over 50 years students have been told that water is roughly $\ce{sp^3}$ hybridized.  The general description is that there are two equivalent O-H sigma bonds and two equivalent lone pairs orbitals.  The lone pair - lone pair repulsion is greater than the sigma bond - sigma bond repulsion, so the hybridization changes as described above and the lone pair-O-lone pair angle opens up slightly and the $\ce{H-O-H}$ angle closes down to the observed 104.5 degrees.
With the advent of photoelectron spectroscopy it was found that the two lone pairs in water were not equivalent (2 signals were observed for the lone pairs).  Now, the hybridization of water is described as follows:

2 $\ce{sp^4}$ O-H sigma bonds
one lone pair in a p orbital
and the second lone pair in an $\ce{sp}$ orbital