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I'm calculating the force of a body hitting water from a high height. So far I've figured out the velocity at the time of collisions would be 40 m/s. The body is 78 kg. I seem to need the time it takes for it to accelerate to a stop after hitting the water but I can't find any concrete numbers for that, what would be a good estimation?

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  • $\begingroup$ How do I find the impulse force without how long it takes for the body to accelerate? Your answer is good but I don't think I explained what I needed properly. The body hitting the water is a person and I'm trying to calculate the force the impact would have on their body. Wouldn't you need time to solve because force is change in momentum over change in time? $\endgroup$ – user108559 Feb 17 '16 at 12:23
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Assuming that the given body is spherical in shape whose density is lesser than that of water and the height of the water pool is so large that the body doesn't hit the bottom before stopping.

After making those assumptions, the question can be easily solved using the viscous and the buoyant forces.

The buoyant force is given by = $V\rho g$ where $V$ is the volume of the body, $\rho$ is the density of water and $g$ is the acceleration due to gravity.

The viscous on the other hand is given by = $6\pi \eta rv$

where $\eta$ is the coefficient of viscosity of water, $r$ is the radius of the sphere (body) and $v$ is the velocity of the body.

The buoyant force always pushes the body up to the surface and the viscous force always acts in a direction that is opposite to the direction of motion. In your case, the force is upwards until the body stops.

By using Newton's laws you can find the net acceleration of the body due to the above two forces as well as force due to gravity. The acceleration is a function of time so you would probably would've to do some integration to obtain the final answer.

Go through the question again and find out what your question expects you to use and what not.

If your question was just asking for the impulse provided to the body after it hits the surface and until it stops, you can calculate it easily using the definition of impulse.

Impulse = Change in momentum

$J = P_{\text{final}} - P_{\text{initial}} = P_{\text{initial}}$ since the body is at rest at the final point.

Hence, calculating impulse would be calculating the initial momentum which is quite easy as the question has already provided the mass and the initial velocity of the body.

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    $\begingroup$ Not overkeen on this derivation as the flow is most probably not laminar? The impulse is derived in this paper but there is no mention of time. iwwwfb.org/Abstracts/iwwwfb15/iwwwfb15_38.pdf $\endgroup$ – Farcher Feb 17 '16 at 10:09
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    $\begingroup$ Indeed. This approach is adequate (or even more than) for game physics but not to be taken seriously. At even moderately high impact speeds the naive drag expression will be badly wrong but surprisingly so with the naive bouyancy expression as the body will change the local pressure environment markedly. $\endgroup$ – dmckee Feb 17 '16 at 15:09
  • $\begingroup$ I doubt that a calculation of viscous forces for water is correct for this problem. The person will be initially moving at 40 m/s, so entry into the water should produce turbulent flow around the person. If you know the drag coefficient for the person, you should be able to set this up as a fluid drag problem, where the drag force decreases as the person's velocity decreases (F is proportional to v^2). This, of course, means that impulse will vary as the person decelerates. $\endgroup$ – David White Jul 25 '16 at 22:51

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