# The maximum velocity of an object in water [closed]

Suppose we have the mass and volume of an object, like a sphere, with density less than water. We grab it under the water and leave it, so it will fly up to the surface. What will be the maximum velocity it reaches to? and what is its acceleration at the beginning and also at a certain time $$t$$?

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• You're looking for the terminal velocity: en.wikipedia.org/wiki/Terminal_velocity – Gert Mar 2 at 19:21
• You need to give values to work with, and also include the depth it's submerged to. – TechDroid Mar 2 at 19:23
• @TechDroid: I don't think he's looking for a numerical solution. – Gert Mar 2 at 20:23
• @Gert. Are you sure the terminal velocity link will help the guy, it's meant for a falling body and he seems to be perplexed with the buoyancy force. – TechDroid Mar 2 at 20:43
• @TechDroid: of course. The terminal velocity applies as much for floating spheres as it does for sinking ones. The only difference is that the gravity and drag vectors are inverted. – Gert Mar 2 at 20:46

Let's say we have a floating sphere, which is released below the surface of a liquid, at $$t=0$$:

[![Force diagram][1]][1]

Three forces act on in while it's floating towards the surface:

1. Buoyancy $$F_B$$:

If the sphere's radius is $$R$$, $$\rho$$ the density of the liquid, then the buoyant force is given by:

$$F_B=\frac43 \pi R^3\rho g$$

1. Weight $$mg$$

2. Drag force $$F_D$$:

The liquid resists motion of the sphere due to viscous friction, resulting in the drag force $$F_D$$.

Depending on velocity, the drag force will be of the form, where $$v$$ is the velocity:

Low velocity: $$F_D \propto v$$

Higher velocities: $$F_D \propto v^2$$

At very low velocities (low Reynolds number), [Stokes' Law][2] can be applied:

$$F_D=6 \pi R \eta v$$

where $$\eta$$ is the viscosity of the liquid.

We'll use it here to illustrate the terminal velocity concept.

The force balance and equation of motion of the sphere is, with $$m$$ the mass of the sphere:

$$ma=F_B-F_D-mg=\frac43 \pi R^3\rho g-6 \pi R \eta v-mg$$

The sphere accelerates upward until:

$$\frac43 \pi R^3\rho g-6 \pi R \eta v-mg=0$$

because then $$ma=0$$ and the sphere no longer accelerates: it has reached terminal velocity $$v_t$$:

$$\frac43 \pi R^3\rho g=6 \pi R \eta v_t+mg$$

From which $$v_t$$ can easily be extracted.

From:

$$ma=\frac43 \pi R^3\rho g-6 \pi R \eta v-mg$$

the evolution of velocity in time, $$v(t)$$, can also be determined:

$$m\frac{\mathrm{d}v(t)}{\mathrm{d}t}=\frac43 \pi R^3\rho g-6 \pi R \eta v(t)-mg$$

from which $$v(t)$$ can be determined by simple integration. If the density of the sphere is $$\rho_S$$ then we can rewrite the above equation as:

$$m\frac{\mathrm{d}v(t)}{\mathrm{d}t}=\frac43 \pi R^3(\rho -\rho_S)g-6 \pi R \eta v(t)$$

Then:

$$\int_0^{v(t)}\mathrm{d}v(t)\frac{m}{\frac43 \pi R^3(\rho -\rho_S)g-6 \pi R \eta v(t)}=\int_0^t\mathrm{d}t$$

$$-\frac{1}{6 \pi R \eta}\ln{\Big[\frac{\frac43 \pi R^3(\rho -\rho_S)g-6 \pi R \eta v(t)}{\frac43 \pi R^3(\rho -\rho_S)g}\Big]}=t$$

$$\frac{\frac43 \pi R^3(\rho -\rho_S)g-6 \pi R \eta v(t)}{\frac43 \pi R^3(\rho -\rho_S)g}=\exp(-6 \pi R \eta t)$$

$$\frac43 \pi R^3(\rho -\rho_S)g-6 \pi R \eta v(t)=\frac43 \pi R^3(\rho -\rho_S)g\exp(-6 \pi R \eta t)$$

From which $$v(t)$$ can be extracted. The velocity tends exponentially to $$v_t$$.

The buoyant force on a submerged object is equal in magnitude and opposite in direction to the force of gravity on the water that it happens to be displacing. If that is greater than the force of gravity on the object, the object will accelerate upward.

However, as you pointed out, there is a potential limit to its acceleration. Water, even more than air, exerts a drag force on anything trying to move in it (unrelated but interesting side-note: that's why sea creatures are more streamlined than land creatures). Like any drag force, it is related to the velocity of the moving object.

So the terminal velocity of a sphere floating up toward the surface of water is the velocity at which the drag force is equal to the buoyant force. Drag can be a bit of a tricky subject, but in this case it shouldn't be too bad. Set this equation equal to the buoyant force and solve for velocity. The drag coefficient for a sphere is about 0.47.

As for the acceleration at a given time t, that's less easy, but still doable if you're familiar with differential equations. Acceleration equals the net force (gravity + buoyancy + drag) divided by the mass, so you'll set up your differential equation such that dv/dt = (-mg + pgV - .5pCAv^2)/m. It's not pretty, but you can solve it with a computer and you end up with the following (this is just copy/pasted out of Mathematica; I used a radius of 1m, and a mass of 1 kg, which puts it far below the buoyancy of water).

eq = DSolve[v'[t] == -98/10 + 9800 4/3 \[Pi] - 1/2 1000 v[t]^2 47/100 \[Pi], v[t], t][[1, 1]] // FullSimplify
D[v, t] /. eq

v[t] -> 7/10 Sqrt[1/141 (4000 - 3/\[Pi])] Tanh[14 Sqrt[47/3 \[Pi] (-3 + 4000 \[Pi] (t - 15 C[1])]
v'[t] -> 49/15 Sqrt[(4000 - 3/\[Pi]) \[Pi] (-3 + 4000 \[Pi])] Sech[14 Sqrt[47/3 \[Pi] (-3 + 4000 \[Pi])] (t - 15 C[1])]^2


Like I said, not pretty.