# Equation to find how far into water you will sink depending on how high you jump from [duplicate]

I am looking for an equation to which I can plug in numbers to figure out to what depth a person will sink into water before momentum is lost and they begin to float upwards, depending on how high they jump from.

To keep it simple, I am assuming 90 degree straight fall, feet first, and the position is kept in the water until momentum is lost.

• I think the harder question is whether you are assuming that they are being slowed by just the buoyant force of the water, and if you do assume drag from the water, whether you also factor in the turbulence of the water. The full, accurate equation will be very non-trivial to solve. – Jerry Schirmer Nov 4 '16 at 5:28
• Why? I can understand you wanting to calculate this for yourself, as an exercise, but asking someone else to calculate for you? If all you want is a realistic answer, there is no substitute for actual data from experiments. (-1 from me for lack of research effort.) – sammy gerbil Nov 4 '16 at 17:22
• @JerrySchirmer If you read through both answers on the duplicate I linked, you'll see that it might not be that complicated actually. Specifically my answer gives a really simple approximation for it that turns out to work astonishingly well. And it is, hands down, my favorite physics approximation ever. – tpg2114 Nov 5 '16 at 4:38

The moment the object is submerged, two forces act on it:

$F_B=V(\rho_{water}-\rho_{object})g$, the buoyancy force and a drag force $F_D$.

$F_D$ is usually modelled as $F_D=kv$ where $v$ is speed and $k$ a drag coefficient.

The equation of motion then is: $$ma=-F_B-F_D$$ $$ma=-F_B-kv$$ $$m\frac{dv}{dt}=-F_B-kv$$ $$-m\frac{dv}{F_B+kv}=dt$$ $$-m\int_{v_0}^{v(t)}\frac{dv}{F_B+kv}=\int_0^tdt$$ Where $v_0$ is the speed just after immersion. If the object falls from height $H$, then $\frac12 mv_0^2\approx mgH$. $$-\frac{m}{k}\ln\frac{kv(t)+F_B}{kv_0+F_B}=t$$ $$kv(t)+F_B=(kv_0+F_B)e^{-\frac{k}{m}t}$$ $$v(t)=\frac1k\Big((kv_0+F_B)e^{-\frac{k}{m}t}-F_B\Big)$$ $$\frac{dx(t)}{dt}=\frac1k\Big((kv_0+F_B)e^{-\frac{k}{m}t}-F_B\Big)\tag{1}$$ $$\int_0^{x(t)}dx(t)=\int_0^t\Big[\frac1k\Big((kv_0+F_B)e^{-\frac{k}{m}t}-F_B\Big)\Big]dt$$ $$x(t)=\frac1k\int_0^t\Big((kv_0+F_B)e^{-\frac{k}{m}t}-F_B\Big)dt$$ $$x(t)=\frac1k\Big[\int_0^t\Big((kv_0+F_B)e^{-\frac{k}{m}t}dt-\int_0^tF_Bdt\Big]$$ $$x(t)=\frac1k\Big[-\frac{m}{k}(kv_0+F_B)\big[e^{-\frac{k}{m}t}\big]_0^t-F_Bt\Big]$$ $$x(t)=-\frac{m}{k^2}(kv_0+F_B)(e^{-\frac{k}{m}t}-1)-\frac{F_Bt}{k}\tag{2}$$ At some point $v(t)$ becomes zero and the buoyancy then causes the object to reverse direction, so there is a minimum value for $x(t)$. We can find $t_{min}$ as follows: $$\frac{dx(t)}{dt}=\frac1k\Big((kv_0+F_B)e^{-\frac{k}{m}t}-F_B\Big)=0$$ $$e^{-\frac{k}{m}t}=\frac{F_B}{kv_0+F_B}\tag{3}$$ $$t_{min}=\frac{m}{k}\ln\frac{kv_0+F_B}{F_B}\tag{4}$$ $$e^{-\frac{k}{m}t}-1=-\frac{kv_0}{kv_0+F_B}\tag{5}$$ $$x_{min}=\frac{mv_0}{k(kv_0+F_B)}-\frac{mF_B}{k^2}\ln\frac{kv_0+F_B}{F_B}\tag{6}$$

$(6)$ represents the deepest an object would sink, before buoyancy starts making it move upwards again.

• Flow will be turbulent, so you should use $F_{drag}=kv^2$ as a first approximation. – Deep Nov 4 '16 at 9:22
• "Flow will be turbulent," You don'r know that a priori. – Gert Nov 4 '16 at 11:24
• Does your formula agree with experience? – sammy gerbil Nov 4 '16 at 17:18
• It's a theoretical treatment, Sammy. Food for thought. – Gert Nov 5 '16 at 2:14
• @Gert Kinematic viscosity of water is $\sim 10^{-6}$. For a jump from a height of say $0.1$ m, entry velocity will be $\sim 1$ m/s. Length scale associated with the body, even if you just consider the feet, is $\sim 0.1$ m. So Reynolds number works out to be $\sim 10^5$. So a priori assumption of turbulent flow is justified. You may drop an object about the size of your palm into water and see for yourself which kind of flow results. – Deep Nov 5 '16 at 5:38

The Federation Internationale de Natation (FINA), the world sporting authority on swimming and diving, uses the following equation (taken from table FR5.3 in Annexe 1.2 of its Facilities Regulations) for the minimum safe depth of water $D$ for High Diving from a height of $H$ metres above the surface :

$$D = 0.143H+3.04$$

This can be taken as an indication of the maximum depth achievable from jumping into water from a given height. The data (column 1 : Platform Height, colmn 2 : Minimum Safe Depth) is

1 ... 3.2
3 ... 3.5
5 ... 3.7
7.5 ... 4.1
10 ... 4.5

The above formula is based on recommendations for dives of up to 10m. However, Annexe 4.2 recommends a safe water depth of 6m for all dives between 5 and 27m. So I would guess that 7m is a fair estimate of the maximum depth you could reach from any height, unless you were carrying additional weight. This is confirmed by Laso Schaller's World Record Jump from 58.8m in August 2015. Remarkably, he plunged only about 4m under the surface.

You should also consider the likelihood of survival. Jumping into water states that a jump of more than 76m is highly likely to be fatal, and that a jump of more than 46m is highly likely to result in spinal injury - unless, like Laso Schaller, you have years of practice.