Answering your questions in reverse order:
Yes, a long pointy object (like your arms over your head, in a dive, or your pointed toes in a feet-first entry) will make a big difference. Remember the tongue-in-cheek adage, "it's not the fall that kills you; it's the sudden stop?" That is exactly what differentiates a fall onto concrete from a fall into water: how sudden is the stop. And making that stop LESS sudden (decreasing the magnitude of deceleration during the stop) is exactly how airbags save your life in a car crash. One can decrease the magnitude of deceleration by reducing the ratio $(\Delta V / \Delta t)$. Since there is roughly a linear relationship between time and distance traveled during the instant of impact, you can achieve the same effect by reducing the ratio $(\Delta V / \Delta s)$ where $s$ = distance traveled during the deceleration event. The easiest way to do this is to lengthen $s$.
One thing to remember about the water fall statistics is that a large number of them are likely "unpracticed". These are not olympic divers working up to 250 feet. A large proportion of them are unconditioned people forced into a water "escape"; or, worse, are people TRYING to die.
Assuming you are doing the right thing, and optimizing your form for water entry, you will simultaneously be minimizing your wind resistance during the fall:
1.) A fall from 30 feet will result in a velocity of roughly 44 ft/s = 30 mph.
2.) A fall from 100 feet will result in a velocity of roughly 80 ft/s = 54 mph.
3.) A fall from 150 feet will result in a velocity of roughly 97 ft/s = 66 mph.
4.) A fall from 250 feet will result in a velocity of roughly 125 ft/s = 85 mph.
The first case is a tower jump I did for the Navy, and is trival for anyone who is HWP and doesn't belly flop. The second is an approximation of a leap from a carrier deck, which the tower jump was supposed to teach you how to survive (be able to swim after the fall). The third is only 20% faster entry speed (and force) and should be survivable by anyone in good shape and able to execute good form (pointed toe entry, knees locked, head up, arms straight up). The La Quebrada cliff divers routinely dive from 125 feet as a tourist attraction. If forced to choose, I'd pick a feet-first entry at 150 feet over a dive at 125.
So the interesting part is the stretch from 150 to 250 feet. My guess is that the limit for someone voluntarily performing repeated water dives/jumps from a height of $x$ will show $x$ to be somewhere around $225 \text{ feet} \pm 25 \text{ feet}$.
EDIT: There are documented cases of people surviving falls from thousands of feet (failed parachute) onto LAND. These freaky cases of surviving terminal velocity falls do not answer the question practically; but they are there.
For example, Vesna Vulović is the world record holder for the biggest surviving fall without a parachute.
If we assume continuity, infinite speed of sound, no viscosity, and laminar flow, then the key is in Bernoulli's equation. In general the pressure variation is very complicated, but to get some ideas how it works we can consider the pressure variation a very simple system that can be calculated analytically.
I came up with this simple system:
Let's say somewhere in the water far from the surfaces, suddenly a cavitation bubble of radius $R$ with pressure $P_0-\Delta P$ is produced. Where $P_0$ is the atmospheric pressure, let's assume that the scale of the system we are playing with is small enough that the pressure of undisturbed water anywhere is equal to the atmospheric pressure. Because of the pressure drop, water will start moving radially to fill the bubble as shown in the picture. If the flow is continue and laminar then we have the following flux relation
$v 4\pi r^2= constant$
$v=\frac{C}{r^2}$..........................................(1)
Now Bernoulli's equation gives
$P_0=P(r)+\frac{1}{2}\rho v^2=P(r)+\frac{1}{2}\rho \frac{C^2}{r^4}$
Because $v=0$ at a point far from the bubble. Boundary condition at $r=R$ gives
$P_0=(P_0-\Delta P)+\frac{1}{2}\rho \frac{C^2}{R^4}$
Eliminating $C$ and $\rho$ we get
$P(r)=P_0+\Delta P \frac{R^4}{r^4}$
As expected $r\rightarrow \infty$, $P(r)\rightarrow P_0$. Thus the pressure change fades away as we go farther from the source of disturbance.
In a more general case where the above assumptions still hold, the velocity profile of the flow equivalent to our eq.(1) is quite complicated. Eq. (1) can be replaced with a general flux equation which holds along a streamline
$vA=constant$..................................(2)
Where $A$ is the cross sectional area of a streamline portion. A typical flow's streamlines caused by a moving object look like this
We can view eq.(2) as an equation that holds in the moving object's frame. As we can see in the picture above, initially the streamlines are uniformly separated. Let's denote the cross sectional area of each slice of stream line portion as $A_0$, from here we know that if the cross sectional area of a stream line portion equals to $A_0$ then its pressure is unchanged or $P_0$. We can see that the streamlines near the object are denser, which means that their cross sectional area are smaller than $A_0$. Thus from eq.(2) we realize that the water is moving faster there, and Bernoulli's equation says that the pressure there is lower than $P_0$. As we move perpendicularly to the flow, away from the moving object the streamlines' cross sectional area get more and more similar to that of undisturbed ones and so does the pressure there. Therefore it can explain how the pressure disturbance decreases over distance from the source.
Best Answer
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.