Force of Body Impact Upon Water

Question

So in 2014 I fell from 70 feet and landed flat on my back on slowly moving water (.5ft/sec) on the Willamette River. I weighed aproximately 190 pounds at the time. I was curious if someone would help me understand the amount of force my body experienced during the moment of inertia. It was enough to cause short-term internal bleeding lasting only about 10 seconds. I coughed up approximately half a pint to a full pint of blood. I didn't receive any medical attention afterward.

I need this answer in a relatively simple format. Most of the terms I am using are what I found from using online calculators but I don't truly understand the terms, the answers, or the equations used to find them

Answer 1

Using the answers from a previous post, let's try things with your specific scenario.

We can use what Count Iblis says here:

The density of the human body is almost the same as that of water, so you would expect that you'll lose most of your velocity after penetrating a depth equal to the width of your body.

Since you landed "flat on your back" your width in this case is the one perpendicular to your entry into the water: roughly how thick you were front-to-back. Let's say $0.2 \ m$.

The next step is to figure out your velocity, $v$, when you hit the water. Because energy is conserved we know that if you fell (say, from a standing position, without running or jumping first) that the potential energy, $PE$ you started with will equal the purely kinetic energy, $KE$ at impact.

Thus

$$ \begin{aligned} KE_{impact} &= PE_{initial} \\ &= mgh \\ &= 86.2 \ kg \cdot 9.8 \ \frac{m}{s^2} \ \cdot 20.3 \ m\\ &= 17148.6 \ \frac{kg \cdot m^2}{s^2} \\ &= 17148.6 \ J \end{aligned} $$

Where $m$ is your mass ($86.2 \ kg$), $h$ is the height you fell from ($20.3 \ m$), and $g$ is the accelleration due to gravity on Earth ($9.8 \ \frac{m}{s^2}$). I add the last line there so you can see the more familiar unit of energy, joules.

Then we can use Newton's impact depth method that gives us a rough approximation of the total force exerted on your body by the time you came to a stop in the water:

$$\begin{align} F &= \frac{KE_{impact}}{w} \\ &= \frac{17148.6 \ \frac{kg \cdot m^2}{s^2}}{0.2 \ m} \\ &= 85743 \ \frac{kg \cdot m}{s^2} \\ &= 85743 \ N \end{align} $$

By this approach we estimate the force exerted on the parts of your body that struck the water to be $~85.7 \ kN$.

Answer 2

For impact the usual unit to express the severity is the G-load. For instance, when cars are crash tested, the impact that the crash test dummies suffer is measured with accelerometers inside the dummies. Airbags can reduce the peak G-load to less than 10 times the load of normal gravity, allowing the passenger to walk away with only bruises.

In another answer to this question SE contributor D. Betchkal already pointed out an answer to a stackexchange question about force of an impact upon water

Stackexchange contributor Floris writes:
There is an additional complication which relates to the shape of the contact area - you may be familiar with the "belly flop", where you fall flat on the water and it hurts a lot. This is not just because you slow down quickly - it is because there is a brief moment when the contact point between your body and the water moves faster than the speed of sound in water, and this results in an "attached shock wave" which can cause the pressure of the water to briefly become very high.

Falling from large height in water the best case scenario is that you enter the water perpendicular, feet first. That way you penetrate deeper, hence the deceleration is spread out over more time, hence the force of deceleration is less.

Second best scenario is what happened to you; your spine was perpendicular parallel to the surface of the water. (Had you hit the water at some angle severe spinal injury was very likely.)

The peak deceleration may have been very, very high, but lasting only a very, very brief instant. A quick calculation may give a reasonable estimate for your average deceleration, but there may well have been a far higher peak deceleration.

It seems to me the best indicator of the peak deceleration you experienced is the fact that you recovered from your injury without any medical treatment. A G-load over 100 G (no matter how short) is considered lethal. 50 G: possibly survivable, but with severe injury.