# Force of an impact on water

Today, at the Brazilian news, I heard that a men fell from a cruise into the sea, from a height of 50m, and decided to see if he could have survived. I took his weight for being 80kg.

I did $V^2=V_0^2 + 2 G H$, this resulted $V = 31.3 m/s$ or $112.69 km/h$. Also $F = m a=784 N$.

How to relate this to the surface tension of water to get the actual strength of the impact? Assuming he fell standing and the area of contact was $700cm^2$, two feet side by side.

• Jul 23 '14 at 15:09

Impact on water is a very complex topic.

Your simple calculation just figures out the velocity of a free-falling body after a 50 m drop. That just tells you the initial relative velocity of body and water surface. It doesn't tell you much about the force at impact, or whether the person survived.

There are two things that might kill on impact: high local pressure on the surface of the body - this can cause lacerations, bleeding, fractures; and a deceleration of the body that causes internal organs (the brain or the kidneys) to rip loose, resulting in severe internal bleeding and death.

The key to survival, then, is to minimize the local pressure and the deceleration. This is why you see stunt men dive in the most streamlined manner possible - they try to minimize the area with which they enter the water, which limits the force on the body and thus maximizes the time to decelerate. It also helps to fall into "less dense" water - for example, the frothy water right behind a cruise ship will have a lot of bubbles in it, and this will increase the distance you move before slowing down.

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 (see for example http://www.dtic.mil/dtic/tr/fulltext/u2/a259783.pdf - a bit of a long paper...).

It seems that the world record for a stunt man surviving a fall into water stands at just over 54 m (Olivier Favre, https://www.youtube.com/watch?v=mLd529gWKJ4). And that is "still" water. As I mentioned above, in frothy water it might be possible to survive a higher drop.

At the same time, people frequently kill themselves by jumping off the Golden Gate bridge - about 70 m height.

Found an interesting link discussing this - the last answer sounds quite credible in its analysis, making some attempt at quantifying the forces during impact. I don't agree with the entire analysis (in particular the "slamming into a stationary body at half the speed" is too much of an approximation for my taste) but it makes some good points: https://wat.lewiscollard.com/archive/www.newton.dep.anl.gov/askasci/gen01/gen01790.htm

A very crude estimate using the impact depth method: 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 (measured in the direction orthogonal to the contact area so if you go in head first, it will be the length of your body). If the width is 30 cm, then an averge stopping force of F would need to be such that F times 30 cm equals the kinetic energy, this yields approximately 170 g for the average decceleration.

Counting just the deacceleration suffered, you can get an idea (the duration of the impact is related to the viscosity of the fluid).

Using the speed you found, the impulse is:

$$J=\Delta p=-(80\times31.3)=-2504\,N\cdot s$$

Estimating a duration for the impact with the water of for example $0.3\, s$, the average force suffered is:

$$F_{average}=\frac{J}{\Delta t}=-8348\,N$$

The deacceleration suffered is:

$$a=\frac{F_{average}}{m}=-104\,m/s^2$$

Or $11\, g.$

Not enough to kill a human if the impact is in a nice position for the vital organs.

• 0.3 seconds for the impact sounds like too long. May 23 at 2:30

Well, what you will have to do is:

First find the force needed to stop the body, taking the initial velocity equal to the velocity when he hits the water. In this case, calculations will be greatly simplified by taking downward acceleration to be zero.

Second, find the surface tension of water (don't know it off of the top of my head, sorry) which is given in $\frac{newtons}{meter}$ Thus find the number of meters needed to stop you.

Finally, calculate the time taken for the body to come to a complete rest, and the net force exerted over the body during that time period and hence the pressure exerted on the body.

Then you would have to find the parameters for death, which should be available on the internet. However, remember that there are many parameters which will be difficult to calculate. You should also remember you are ignoring air resistance.