# Can a human body supercavitate to survive water impact?

Inspired by this analysis of a human (OK, Captain America) hitting water feet first at terminal velocity, I'm wondering if supercavitation would be possible and whether it would improve your chances of survival?

Would a human body reach a high enough speed to supercavitate on impact? Would this reduce the g-force to a survivable level (eg by allowing deeper penetration into the water)? And crucially, when the supercavitation collapses do you still have to deal with an unsurvivable acceleration?

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The problem with supercavitation is that it occurs at a pressure drop, not an increase. The thing that kills you on impact is the pressure in front of you - the water that cannot get out of the way fast enough. Skin drag on the body happens later - when more of the body is already submerged. Most likely you are dead by then...

Let me try (with my rusty hydrodynamics) to estimate a few things.

For a sphere moving through water, the apparent inertia experienced in laminar flow (because when you move the sphere, you move water) is the mass of the sphere plus half the mass of the displaced liquid (this is called the added mass). Of course a human is not even close to a spherical shape, but this gives us an order of magnitude of the forces we are looking for.

For a human with a density approximately that of water, the effect of the added mass can be thought of as hitting another human half their weight (before accounting for skin drag) when they hit the water. You don't hit that human all at once, of course, since your body isn't hitting the water all at once - but at a first approximation, the time it takes to hit that body is your velocity divided by your length, and by conservation of momentum your velocity must decrease by 1/3 in that time. If the density of the water is lower, this ratio will become smaller, and in general is

\begin{align}ratio &= 1-\frac{\rho_1}{\rho_1 + 0.5 * \rho_2}\\ &= \frac{\rho_2}{2\rho_1+\rho_2}\end{align}

Where $\rho_1$ is the density of the body, and $\rho_2$ is the density of the water. From this you can see that if $\rho_2$ becomes $<<\rho_1$, the ratio becomes very small - and there is less initial deceleration. But let's keep going with regular water first.

If you have length $l$ and velocity $v$, the time during which you decelerate will be approximately $l/v$ and in that time your velocity decreases by $v/3$ and momentum decreases by $mv/3$ for a mass $m$. This means an average force

\begin{align}F &= \frac{mv}{3\frac{l}{v}}\\ &= \frac{mv^2}{3l}\end{align}

Putting terminal velocity at 56 m/s (from Wolfram Alpha), length of 2 m (with arms extended above head), and a mass of 70 kg, you get an average force of about 36 kN for the 1/20th of a second that it takes to submerge - that's like an elephant sitting on you for a single frame of a video. Enough to do some real damage to your internal organs (losing 1/3 of your velocity in that time is roughly equivalent to 40 g of deceleration).

You need to lower the density of the water before you hit it. I suppose if you had a machine gun on you and shot a rapid burst at the water just before impact, two good things would happen: you would slow yourself down (because of the recoil of the gun - bullets accelerate down = you accelerate up), and you might be able to create cavitation in the water where you land. This lowers the effective density of the water, meaning that you will decelerate more slowly and have a slightly better chance of surviving that initial hit.

As an aside - the current record for a free fall into water stands at 54 m, which puts the velocity at about 30 m/s (assuming some drag) and the initial deceleration at 20 g. That's about double what fighter pilots get, but it's just about believable that a stunt man can survive that on a good day.

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