Question

terminal velocity derivation
for +1 ​

Answer 1

Answer:

Terminal velocity is defined as the highest velocity attained by an object that is falling through a fluid. It is observed when the sum of drag force and buoyancy is equal to the downward gravity force that is acting on the object. The acceleration of the object is zero as the net force acting on the object is zero.

Answer 2

Explanation:

Where,

vt: terminal velocity

m: mass of the falling object

g: acceleration due to gravity

Cd: drag coefficient

: density of the fluid through which the object is falling

A: area projected by the object

Terminal velocity derivation

Deriving terminal velocity using mathematical terms according to the drag equation as follows:

F=bv2 (drag force)

Where,

b: constant depending on the type of drag

∑F=ma (free fall of an object)

mg−bv2=ma (assuming that the free fall is happening in positive direction)

mg−bv2=mdvdt 1mdt=dvmg−bv2 (differential form of the equations)

∫1mdt=∫dvmg−bv2 (integrating the equations)

∫dvmg−bv2=1b∫dvα2−v2

Where,

α=mgb−−−√ dv=αsech2(Θ)dΘ (after substituting for v=αtanh(Θ))

v2=α2tanh2(Θ)

After integration

1b∫αsech2(Θ)dΘα2−α2tanh2(Θ) 1b∫αsech2(Θ)dΘα2(1−tan2Θ) 1b∫αsech2(Θ)dΘα2sech2(Θ)=1αb∫dΘ=1αbarctanh(vα)+C (using the identity 1−tanh2(Θ)=sech2(Θ))

1mt=1αbarctanh(vα)+C (from original equation)

v(t)=αtanh(αbmt+arctanh(v0α))

By substituting for α=mgb−−−√ v(t)=αtanh(tbgm−−√+arctanh(v0α)) v(t)=mgb−−−√tanh(tbgm−−√+arctanh(v0bmg−−−√))

After substituting for vt

limt→∞v(t)=limt→∞(mgb−−−√tanh(tbgm−−√+arctanh(v0bmg−−−√)))=mgb−−−√ ∴vt=2mgρACd−−−−√