Slowing toward terminal velocity

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Slowing toward terminal velocity

Postby Himself » Fri Jun 27, 2014 4:30 am UTC

I am doing some work involving objects in freefall and am trying to find an equation for how an object would slow after entering a medium where it has a much lower terminal velocity, specifically water. Say, for example, I have an object with a terminal velocity in water of 2 m/s striking the surface at 60 m/s. How could I calculate how much the object would slow down as a function of depth?
I don't think the objects landed in deep water, but I want to cover all bases.
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Re: Slowing toward terminal velocity

Postby Neil_Boekend » Fri Jun 27, 2014 1:52 pm UTC

Short answer: really fast.

There are 3 main parts to this:
1. Difference between the force through the mass of the object (as it is attracted by earth) and the total drag force.
2. Buoyancy. This can either slow down or speed up the decent.
3. Surface tension of the water. This gives an initial deceleration that is even higher than the normal one (a couple of ms later).

1. The force through mass is easy: 9.81*m (in kg). The force through drag is a bit more difficult, but if you have calculated the terminal velocity then you should have the required values for that.
p = 1000 kg/m3
A = projected area, in m3
Cd = drag coefficent (for example 0.47 for a sphere)

If I assume a 1 meter diameter sphere of 100 kg this becomes:
Force through mass = 981 N
0.5*1000*602*0.75*0.47= 63.45 KN.
Thus the force upwards would be approximately 62,469 N.
a=F/m=62,469/100= 624.69=> so this object would decelerate with approximately 625m/s2 or approximately 64g

2. Buoyancy. Quite easy, subtract the mass of object from the water the object displaces (volume*1000kg/m3) and multiply by g.
For our 1 m3 sphere this is:
290 kg of displaced water and 100 kg of ball. That means there is 290-100*9.81=1908 N pushing it up. This gives an additional deceleration of 1908/100 = 19.08 m/s2 or approximately 2g.

3. Surface tension. I have no idea how to calculate this.
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