If I jump from an airplane straight positioned upright into the ocean, why is it the same as jumping straight on the ground?
Water is a liquid as opposed to the ground, so I would expect that by plunging straight in the water, I would enter it aerodynamically and then be slowed in the water.
When you would enter the water, you need to "get the water out of the way". Say you need to get 50 liters of water out of the way. In a very short time you need to move this water by a few centimeters. That means the water needs to be accelerated in this short time first, and accelerating 50 kg of matter with your own body in this very short time will deform your body, no matter whether the matter is solid, liquid, or gas.
The interesting part is, it does not matter how you enter the water—it is not really relevant (regarding being fatal) in which position you enter the water at a high velocity. And you will be slowing your speed in the water, but too quickly for your body to keep up with the forces from different parts of your body being decelerated at different times.
Basically I'm making a very rough estimate whether it would kill, only taking into account one factor, that the water needs to be moved away. And conclude it will still kill, so I do not even try to find all the other ways it would kill.
Update - revised:
One of the effects left out for the estimate is the surface tension.
It seems to not cause a relevant part of the forces - the contribution exists, but is negligibly small. That is depending on the size of the object that is entering the water - for a small object, it would be different.
(see answers of How much of the forces when entering water is related to surface tension?)
Let's look at this another way: you're just moving from one fluid to another. Sounds harmless, right? By specification of the problem, we're at terminal velocity when we hit the water. The force of drag (in both mediums) is roughly:
$$ F_D\, =\, \tfrac12\, \rho\, v^2\, C_D\, A = \rho \left( \frac{1}{2} v^2 C_D A \right) $$
You can imagine that everything except for the density term is the same as you initially transition from the air medium to water. This isn't perfectly accurate, because these are very different Reynolds numbers, but it's good enough for here.
That means that the force (and correspondingly, acceleration) will simply change by the same factor that the density changes by. Also, we know the original acceleration due to drag was 1g, in order to perfectly counteract gravity, which is the definition of terminal velocity. That leads to a simple estimation of the acceleration upon hitting the water. I'll assume we're at sea level.
$$ \frac{a_2}{a_1} = \frac{ a_2 }{1 g}= \frac{ \rho_{H20} } { \rho_{Air} } = \frac{1000}{1.3} \\ a_2 \approx 770 g $$
The maximum acceleration a person can tolerate depends on the duration of the acceleration, but there is an upper limit that you will not tolerate (without death) for any amount of time. You can see from literature on this subject, NASA's graphs don't even bother going above 100g.
Note that a graceful diver's entry will not help you - that's because an aerodynamic position also increases the velocity at which you hit.
Consider jumping into a swimming pool. Do a barrel-roll (sorry I mean cannon ball, that just kind of slipped out). It's fun, you enter the water nicely and make a huge splash, probably soaking your sister in the process (that'll learn her). Now do a belly flop. Not as fun. You displace exactly the same amount of water in the same time, but this time there is a lot more pain and you come away with red skin and maybe some bruising. The difference? You cover more area in a belly-flop than a cannon ball.
At extreme velocities, accelerating your body's mass of water will kill you anyway. However, what actually kills you is hitting the surface. Dip your hand in water... easy. Now slap the surface.... it's like hitting the table (almost). Pressures caused by breaking the surface make water act more solid on shorter timescales, which is why they say hitting water at high speeds is like hitting concrete; on those short times, it is actually like concrete!
The ocean surface is not as hard as the ground but if you drop from a plane, you would hit it with such a high velocity that the pressure would most likely kill you or cause very serious damage.
Considering air resistance, the terminal velocity of a human, right before reaching the water, would be at most some $150\text{ m/s}$.
If you weigh $70\text{ kg}$, that would amount to a Kinetic Energy of
$$\frac12mv^2=0.5\times70\times150^2\text{ J}=787\ 500\text{ J}$$
Which is a LOT of energy, enough to crush many parts of your body even if you land on water. As ratchet streak mentioned, the water molecules can't move out of the way like they would do if you had fallen from a smaller height because of the high velocity. So you basically hit a semi-solid surface and all that energy comes back to you as a Reaction (Normal) Force.
I'm not a physicist. So I am treading very carefully attempting to answer a question here... :)
A physical example that may help explain this is rock skipping. When you skip a rock, it will 'bounce' off of the water when at high speeds. Eventually it slows enough to no longer bounce but 'sink' into the water.
Picture your body doing the same thing. Your body will not want to sink into the water when going at that initial high speed as your body simply can't displace that water fast enough. So there's a force that acts back upon your body.
For a rock, not a huge deal. For a sack of living meat and blood and brains, it won't look pretty.
When you go fast enough, the water molecules just can't move out of the way fast enough for a soft landing.
