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

  • $\begingroup$ Though that's an intense story (and I'm glad you're alright), this question has already been answered on this site in a nearly identical form. There are several different styles of answers that may be helpful: physics.stackexchange.com/questions/120036/… $\endgroup$ Mar 25, 2018 at 3:38
  • $\begingroup$ I recognize that links have been published prior to this that help explain the force but I'm interested in understanding the information for my own experience. I don't understand the use of equations well enough to calculate it myself. Do you think you could help me understand how to use the equations using body/water density, the angle of impact/ surface area, and mass to find out what the peak of g-force i experienced was... i know its a complex question, thats why i need some help to figure it out. $\endgroup$ Mar 25, 2018 at 4:45
  • $\begingroup$ or maybe I'm just not aware enough of the equations to adapt it. if the equation would give me the answer I would need assistance to learn how to use it $\endgroup$ Mar 25, 2018 at 4:46
  • $\begingroup$ Please pardon my persistance but i figured this may be a potential way to be educated on physiscs using an example that would be easy for me to understand. $\endgroup$ Mar 25, 2018 at 4:50

2 Answers 2


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.


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

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 $~85.7 \ kN$.

  • $\begingroup$ I found an article that says an estimated conversion from newton to g-force suggests that every 784 newtons are equivalent to 1 g. When converting your answer it suggests that I may have experienced somewhere around 109 units of g-force upon impact. Considering that the water would be less dense from predetermined motion and that I impacted at an angle (not flush) the g-force would have been reduced the amount of force I would have experienced had the water been still and I landed at 180 degrees. Does this seem like a realistic theory to how I could've survived? $\endgroup$ Mar 25, 2018 at 19:03
  • $\begingroup$ is there information available to factor in the density of the moving water to give a more accurate measure of the g-force? The article below suggests that I would not have been able to survive the impact. I believe the average movement of the river is 0.5 ft/sec $\endgroup$ Mar 25, 2018 at 19:06
  • $\begingroup$ In all reality, I'm not sure if I hit the water flush or at an angle. $\endgroup$ Mar 25, 2018 at 19:24
  • $\begingroup$ @GarrettFarley to be honest, I'm not sure how to account for the movement of the water, or if it's a factor at all. Also, $1 G$ for a mass of $86 \ kg = 9.8 \cdot 86 \approx 845 N$, so given horizontal entry to the water, I calculate a slightly smaller $102 G$. Compare this to entry vertical to the water, where (if you're an average height of $1.8 \ m$), your total G-force is much smaller - only $11 G$, so your orientation with respect to the water at impact plays a huge role in survival. Remember that my answer is a crude approximation - there may be a more subtle reason why you survived. $\endgroup$ Mar 25, 2018 at 19:41
  • $\begingroup$ I think the primary reason for my survival may have been equal diffusion of pressure through the surface area of my body. I saw a story of a man who survived being thrown by a tornado because he was knocked unconscious by a piece of debris before being lifted. Upon impact, his inability to react resulted in his core structure being able to absorb the brunt of the force of the impact versus the jerking that may have been caused by impacting upon his arms or legs. I believe that this could be the result of a similar occurrence $\endgroup$ Mar 26, 2018 at 3:45

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.

  • $\begingroup$ thank you. I had had trouble finding information on the amount of g-force a body could sustain. $\endgroup$ Mar 25, 2018 at 19:07

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