The forces slowing you are (1) drag as you note and (2) buoyancy. The former, assuming ram drag is the main one, is given by:
$$F_D = -\frac{1}{2}\,A\,\rho_W\,C_D\,v^2$$
where $\rho_W$ is the density of water, $v$ the velocity of the dragged object, $A$ the cross-sectional area presented to the water as you fall and $C_D$ is a fudge factor called the drag co-efficient. $C_D$ is highly dependent on the object's shape and orientation relative to its velocity through the water. To understand more about ram pressure, see my answer here. So you will need to "calibrate" $C_D$ with an observed depth.
The buoyancy force is the weight of the water you displace. So if your density is $\rho_B$ and your mass $m$, then the buoyancy force is $-\frac{\rho_W}{\rho_B}\,m\,g$ (downwards positive).
At last we have your weight $+m\,g$. Therefore, Newton's second law becomes the following differential equation for velocity $v(t)$ (downwards direction positive)
$$m\,\mathrm{d}_t\,v(t) = m\,g\,\left(1-\frac{\rho_W}{\rho_B}\right) - \frac{1}{2}\,A\,\rho_W\,C_D\,v(t)^2$$
we convert this to an equation for velocity $v$ as a function of depth penetrated $y$ by the identity $\mathrm{d}_t\,v = v\,\mathrm{d}_y\,v = \frac{1}{2}\,\mathrm{d}_y\,v^2$, so we are left with:
$$\mathrm{d}_y\,v^2 = 2\,g\,\left(1-\frac{\rho_W}{\rho_B}\right) - \frac{\rho_W}{m}\,A\,C_D\,v^2=- \frac{\rho_W}{m}\,C_D\,A\,\left(v^2+2\,\frac{m\,g}{C_D\,A}\left(\frac{1}{\rho_B}-\frac{1}{\rho_W}\right)\,\right)$$
whence:
$$\log\left(v^2+2\,\frac{m\,g}{C_D\,A}\left(\frac{1}{\rho_B}-\frac{1}{\rho_W}\right)\,\right) = - \frac{\rho_W}{m}\,C_D\,A\,y + C_I$$
where $C_I$ is an integration constant we must now find. As you know, we have $\frac{1}{2} v(0)^2 = g\,h$, where $v(0)$ is your velocity as you hit the water and $h$ the distance you dive from. If in the above equation we measure $y$ downwards from the water's surface, we have:
$$\log\left(2\,g\,h+2\,\frac{m\,g}{C_D\,A}\left(\frac{1}{\rho_B}-\frac{1}{\rho_W}\right)\,\right) = C_I$$
and so we can now work the integration constant out to find:
$$\log\left(\frac{v^2+2\,\frac{m\,g}{C_D\,A}\left(\frac{1}{\rho_B}-\frac{1}{\rho_W}\right)}{2\,g\,h+2\,\frac{m\,g}{C_D\,A}\left(\frac{1}{\rho_B}-\frac{1}{\rho_W}\right)}\right) = - \frac{\rho_W}{m}\,C_D\,A\,y$$
and then find the $y$ that makes $v=0$. So at last we have the description of your penetration depth $d$; it is:
$$d=\frac{m}{\rho_W\,C_D\,A}\,\log\left(1+\frac{h\,C_D\,A}{m\,\left(\frac{1}{\rho_B}-\frac{1}{\rho_W}\right)}\right)$$
Notice how this quantity is negative if your density is greater than that of the water, describing a situation where there were water above the actual surface. This means, of course, that you keep on sinking if you're not buoyant enough.
For very small drags ($C_D\,A\to0$), the above equation becomes:
$$d\approx\frac{h}{\frac{\rho_W}{\rho_B}-1}$$
but I'm almost certain that this will greatly overestimate your penetration depth: it says that your penetration depth will be much deeper than the dive tower is tall.
So you need at least one $d$ observation to work out the value of the unknown $C_D\,A$ - the "fudge factored" effective cross sectional area you present to the water. $C_D$ values for long thin objects are typically about 1. If your cross sectional area (cut through the anatomist's transverse plane) is $0.5\times 0.3=0.15{\rm m^2}$, your mass $90{\rm kg}$, your density with your breath drawn in is $950{\rm kg\,m^{-3}}$ and your drag co-efficient is $1$, then we get, for $d$ and $h$ measured in metres:
$$d=0.6\,\log\left(1+32\,h\right)$$
yielding $d=2.09{\rm m}$ for $h=1{\rm m}$, $d=2.74{\rm m}$ for $h=3{\rm m}$, $d=3.04{\rm m}$ for $h=5{\rm m}$, $d=3.28{\rm m}$ for $h=7.5{\rm m}$ and $d=3.46{\rm m}$ for $h=10{\rm m}$. These don't seem far off what one observes. These will be underestimates because I didn't correctly describe the "transition epoch" where your body is only partly steeped in the water, and therefore the buoyancy in particular is overestimated.
Moreover, surprisingly, these estimates are not far off tpg2114's answer. Certainly, $d$ is a very weak function of $h$ once $h$ rises above $1{\rm m}$, in keeping with the other answer.
Update: Accounting for the "Transition Epoch"
If we account for the stage where the body is entering the water and model the variable buoyancy as being proportional to the length of body steeped in the water, our basic differential equation becomes:
$$\mathrm{d}_y\,v^2 = 2\,g\,\left(1-\frac{\rho_W\,y}{\rho_B\,L}\right) - \frac{\rho_W}{m}\,A\,C_D\,v^2$$
whose solution (subject to the initial value $v(0)^2 = 2\,g\,h$) is:
$$v(y)^2 = \frac{2\, m\,g\, (A \,C_D\, (L\, \rho_B-\rho_W\, y)+m)}{A^2
\,C_D^2\, L\, \rho_B \,\rho_W}-\frac{2\, g\, \left(-A^2\, C_D^2\, h\, L \,\rho_B\,
\rho_W+A\, C_D\, L\, m\, \rho_B+m^2\right)}{A^2
\,C_D^2\, L \,\rho_B\, \rho_W}\,\exp\left(-\frac{A
\,C_D \,\rho_W\, y}{m}\right)$$
and when the body is fully steeped ($y=L$) the squared velocity is:
$$v(L)^2 = \frac{2\, g\,\left(m\, (A\, C_D\, L\, (\rho_B-\rho_W)+m)-\left(A\, C_D \,L \,\rho_B \,(m-A\, C_D\, h \,\rho_W)+m^2\right)\exp\left(-\frac{A\, C_D\, L \,\rho_W}{m}\right)\right)}{A^2\, C_D^2\, L \,\rho_B \,\rho_W}$$
so the depth of penetration beyond the body's length is given by the equation:
$$d=\frac{m}{\rho_W\,C_D\,A}\,\log\left(1+\frac{h_{eff}\,C_D\,A}{m\,\left(\frac{1}{\rho_B}-\frac{1}{\rho_W}\right)}\right)$$
where now the quantity $h_{eff}$ is given by:
$$h_{eff}=\frac{m\, (A\, C_D\, L\, (\rho_B-\rho_W)+m)-\left(A\, C_D \,L \,\rho_B \,(m-A\, C_D\, h \,\rho_W)+m^2\right)\exp\left(-\frac{A\, C_D\, L \,\rho_W}{m}\right)}{A^2\, C_D^2\, L \,\rho_B \,\rho_W}$$
So now we calculate $h_{eff}$ for the data above ($A=0.15{\rm m^2}$, $m=90{\rm kg}$, $\rho_B=950{\rm kg\,m^{-3}}$, $C_D=1$ and assuming $L=1.9{\rm m}$) with the diving heights of 1, 3, 5, 7.5 and 10 metres:
$$\begin{array}{ll}h=1{\rm m}&h_{eff} = 0.176319{\rm m}\\h=3{\rm m}&h_{eff} = 0.260607{\rm m}\\h=5{\rm m}&h_{eff} = 0.344895{\rm m}\\h=7.5{\rm m}&h_{eff} = 0.450254{\rm m}\\h=10{\rm m}&h_{eff} = 0.555614{\rm m}\end{array}$$
and so, when we put these values into $d=\frac{m}{\rho_W\,C_D\,A}\,\log\left(1+\frac{h_{eff}\,C_D\,A}{m\,\left(\frac{1}{\rho_B}-\frac{1}{\rho_W}\right)}\right)$ we get:
$$\begin{array}{ll}h=1{\rm m}&d = 1.13073{\rm m}\\h=3{\rm m}&d = 1.33494{\rm m}\\h=5{\rm m}&d = 1.48701{\rm m}\\h=7.5{\rm m}&d = 1.63506{\rm m}\\h=10{\rm m}&d = 1.75372{\rm m}\end{array}$$
giving the total depths of penetration of your feet (the above values plus $1.9{\rm m})$:
$$\begin{array}{ll}h=1{\rm m}&d = 3.03{\rm m}\\h=3{\rm m}&d = 3.23{\rm m}\\h=5{\rm m}&d = 3.39{\rm m}\\h=7.5{\rm m}&d = 3.54{\rm m}\\h=10{\rm m}&d = 3.65{\rm m}\end{array}$$
as you can see, a reasonable accounting for the transition epoch adds quite a bit of depth for shallow dives (a whole metre for a 1m dive) but only 20cm for the 10m dive.
On entering the water, the acceleration throughout the transition epoch is:
$$a(y)=e^{-\frac{A\, C_D\, \rho_W\, y}{m}} \left(-\frac{A\, C_D \, g\, h
\rho_W}{m}+\frac{g\, m}{A\, C_D \, L\, \rho_B}+g\right)-\frac{g\, m}{A\, C_D\, L\, \rho_B}$$
which is maximum at $y=0$ and given by:
$$g-\frac{A\, C_D\, g\, h\, \rho_W}{m}$$
working out to be about $-15\,g$ for a 10m dive, but only $-0.8\,g$ for the 1m dive.
