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.
You're looking at the net force acting on him at the instant he begins falling. And you are correct that, in the absence of air resistance, the net force that is acting on the man the instant after he begins falling in the same, given by $F_g=mg$. However, there's an old saying that says, "it's not the fall that kills you; it's the sudden stop at the end." This couldn't be more true, so to understand the difference between the two falls, we need to look at the force acting on the man as he's hitting the ground.
One good approximation to look at this is through the idea of impulse. Impulse is defined as the change in momentum of an object. At the instant before the man strikes the ground in each scenario, he has some momentum, $p$, given by $p=mv$ where $m$ is the mass of the man (in your example, 50 kg) and $v$ in the instantaneous velocity of the man right before he strikes the ground. At this point I'll do a little bit of hand-waving, and just say that the velocity right before striking the ground will increase as the height increases. Therefore, the $v$ before striking the ground for the 50m drop will be much higher than the $v$ from the 2m height. For the purpose of this problem, exact velocities are not important.
Ok, so now let's have the guy actually hit the ground. Although you may think that this occurs instantaneously, it actually occurs over a non-zero time interval. This is easily seen if you ever watch a collision through the aid of slow-motion. Anyway, so when the guy hits the ground, it takes some non-zero time interval, $\Delta t$ to come to rest. Assuming the guy is hitting the same surface in both scenarios, we'll assume $\Delta t$ is a constant across our trials. (This isn't quite true, but slight variations in $\Delta t$ across drop heights are negligible compared to wildly different impact velocities.)
When the guy hit's the ground, his momentum quickly goes from $p$, which is said is equivalent to the expression $mv$ to zero. This is because after he's fully collided with the ground, he's not moving anymore so his velocity, and therefore his momentum is zero.
There's an equation that can be used to express this change in momentum, or impulse, in terms of force, and it's given by $$F\Delta t=m\Delta v$$ If the final velocity is zero, then $\Delta v$ is just given by $-v$, where $v$ is the velocity just prior to striking the ground. Solved for the impact force, $F$, we get that $$F=\frac{mv}{\Delta t}$$
Since the mass of the man, $m$, is the same no matter what height he is dropped from, and I said that we can (for our purposes) treat the time interval $\Delta t$ as a constant as well, you can see that the impact force, $F$, is proportional to the velocity $v$ the jumper has when he hits the ground. And as a mentioned earlier, higher drop height means higher final velocity, and therefore larger impact force.
Best Answer
The normal force from the ground must be large enough to give you an upward acceleration, that is large enough so that you don't melt into the ground. First of all, this means that the normal force will be time dependent: while you are in the air, the normal force is zero; then, as you hit the ground, the normal force grows very rapidly, causing acceleration so that you slow down; finally, the normal force reduces again so that when your velocity is zero, the normal force is equal to you weight, so that you remain on the ground.
When you are jumping from 1 meter, your velocity when you hit the ground is small, and the acceleration needed to make you stand still is small. Hence the maximum normal force does not need to be extremely large.
When you are jumping from 20 meters, your velocity when you hit the ground is large, and the acceleration needed to make you stand still is large. Hence the maximum normal force needs to be large*.
*Actually, the impulse must be large, which is the time integral of the normal force over the period of landing, from first contact to stand-still. This impulse is equal to the change in momentum.