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The maximum velocity which an object can attain as it falls through a fluid (air is the most common example) is known as terminal velocity. It occurs when the sum of the buoyancy and the drag force from the fluid is equal to the downward gravitational force that is acting on an object. In these cases, the net force acting on the object is zero, which is why the object has zero acceleration.

In fluid dynamics, it is said that an object is moving at terminal velocity if the object’s speed is constant due to a restraining force exerted on the object by the fluid in which it is moving. The terminal velocity of an object can be changed by properties of the fluid, the mass of the object, and the object’s projected cross-sectional surface area.

In this article, students will learn how to derive an expression for terminal velocity.

Using the mathematical terms, terminal velocity formula derivation without considering buoyancy effects is given below.

\[Vt = \sqrt{\frac{2mg}{pACd}}\]

Where,

Vt represents the terminal velocity.

mis the mass of the falling object.

g is the acceleration of the object due to velocity.

Cd is the drag coefficient.

p is the density of the fluid through which the object is falling.

A is the projected area of the object.

It is considered that in reality, an object can attain terminal velocity asymptotically.

The buoyancy effect is the upward force that is acting on the object from the surrounding field. The buoyancy effect can be considered in the formula using the Archimedes Principle. Here, the mass m has to be reduced by the displaced fluid’s mass, which is pV, with V being the volume of the object. After this take Mr = m - pV instead of m in all the subsequent formulas.

The following solution can be used to answer the question ‘How to derive the expression for terminal velocity?’.

Terminal velocity derivation using the mathematical terms as per the drag equation is given below.

\[F = bv^{2}\] (drag force)

Where,

b: constant depending on the drag types

\[\Sigma F = ma\] (free fall of an object)

\[mg - bv^{2} = ma\] (assuming that the free fall is taking place in a positive direction)

\[mg - bv^{2} = m(dv/dt) . (1/m)dt = dv/(mg - bv^{2})\]

(differential form of the equations)

\[\int (1/m)dt = \int dv/ (mg - bv^{2})\] (integrating the equations)

\[\int dv/ (mg - bv^{2}) = 1/b \int dv/(\alpha^{2} - v^{2})\]

Where,

\[\alpha = \sqrt{mg/b} dv = \alpha sec h^{2} (\theta) d\theta\]

(after substituting for \[v = \alpha tan h (\theta)\] )

\[v^{2} = \alpha^{2} tan h^{2} (\theta)\]

After integration,

\[1/b \int [\alpha sec h^{2} (\theta) d \theta]/[\alpha^{2} tec h^{2} (\theta)] = 1/b \int [\alpha sec h^{2} (\theta) d \theta]/[\alpha^{2} (1 - tan^{2} \theta)]\]

\[1/b \int [\alpha^{2} sec h^{2} (\theta) ] = 1/ab \int d\theta\] (using \[= 1/ab [arc tan h(v/a) + C\] (the identity) \[1 - tan h^{2} (\theta) = sec h^{2} (\theta)\])

\[(1/m) t = (1/ab) arc tan h(v/a) + C\] (from original equation)

\[v(t) = \alpha tan h[(ab/m) t + arc tan h(v_{0}/\alpha)]\]

By substituting for \[\alpha = \sqrt{mg/b}\],

\[v(t) = \alpha tan h[t\sqrt{bg/m} + arc tan h(v_{0}/\alpha)]\]

\[v(t) = \sqrt{mg/b} tan h[t\sqrt{bg/m} + arc tan h(v_{0} \sqrt{b/mg})]\]

After substituting for Vt,

\[\lim_{t\rightarrow \infty} v(t) = \lim_{t \rightarrow \infty} [\sqrt{mg/b} tan h (t \sqrt{bg/m}) + arc tan h(v_{0} \sqrt{b/mg})] = \sqrt{mg/b}\]

∴ \[Vt = \sqrt{2mg/pACd}\]

Therefore, this is the derivative expression for terminal velocity.

When the buoyancy effects are considered then an object which is falling through a fluid under its own weight can reach a terminal velocity if the net force acting on the object becomes zero. When the object has attained terminal velocity, the weight of the object gets balanced by upward buoyancy force and drag force, which is given by,

W = Fb + D

Where,

W = Weight of the object

Fb = Buoyancy force, and

D = Drag force acting on the object.

In case the object falling is spherical in shape, then the expression for the three forces (W, Fb, D) are given below.

\[W = (\pi /6) d^{3} PsG\],

\[Fb = (\pi/6) d^{3} PG\],

\[D = Cd (1/2) pV^{2} A\]

Where,

d is the diameter of the spherical object.

G is the gravitational acceleration.

P is the density of the fluid.

Ps is the density of the object.

\[A = (1/4) \pi d^{2}\] is the projected area of the sphere.

Cd is the drag coefficient.

V can be considered as characteristic velocity (taken as terminal velocity Vt).

Substitution of equations (2 - 4) in equation (1) can derive an expression for terminal velocity, which is as follows:

\[Vt = \sqrt{[4gd/3Cd\ [Ps - P)/P]}\]

FAQ (Frequently Asked Questions)

1. Can There be any Increase or Decrease in the Terminal Velocity?

Answer: Terminal velocity can be termed as the speed at which your drag force equals your body weight. So according to this theory, it seems that it is possible to increase or decrease the terminal velocity by making some changes in your weight or shape, or altitude. Thus if the shape of the object changes (e.g., by opening a parachute), its drag force will increase causing a decrease in the terminal speed. So, if a skydiver is falling faster than the terminal speed, his speed decreases when he opens the parachute.

2. Does Terminal Velocity Exist in a Vacuum?

Answer: In a vacuum, there is no terminal velocity because terminal velocity is attained when the force of gravity gets perfectly equal to the forces of friction and the air resistance. If we eliminate the air resistance and the force of friction from the atmosphere, then there is nothing to oppose the fall of an object; thus, it will fall faster and faster. Hence it is proved that terminal velocity does not exist in a vacuum rather it can take place in fluid (for example air).