Editing Projectile motion (section)

=== Trajectory of a projectile with Stokes drag ===
Stokes drag, where <math>\mathbf{F_{air}} \propto \mathbf{v}</math>, only applies at very low speed in air, and is thus not the typical case for projectiles. However, the linear dependence of <math>F_\mathrm{air}</math> on <math>v</math> causes a very simple differential equation of motion
:<math>\frac{\mathrm{d}}{\mathrm{d}t}\begin{pmatrix}v_x \\ v_y\end{pmatrix} = \begin{pmatrix}-\mu\,v_x \\ -g-\mu\,v_y\end{pmatrix}</math>
in which the 2 cartesian components become completely independent, and it is thus easier to solve.<ref name="AryaArya1997">{{cite book|author1=Atam P. Arya|author2=Atam Parkash Arya|title=Introduction to Classical Mechanics|url=https://books.google.com/books?id=LxWMAAAACAAJ|date=September 1997|publisher=Prentice Hall Internat.|isbn=978-0-13-906686-3|page=227}}</ref> Here, <math>v_0</math>,<math>v_x</math> and <math>v_y</math> will be used to denote the initial velocity, the velocity along the direction of <var>x</var> and the velocity along the direction of <var>y</var>, respectively. The mass of the projectile will be denoted by <var>m</var>, and <math>\mu:=k/m</math>. For the derivation only the case where <math display="inline">0^o \le \theta \le 180^o </math> is considered. Again, the projectile is fired from the origin (0,0).

{{Collapse top|title=Derivation of horizontal position}}
The relationships that represent the motion of the particle are derived by [[Newton's second law]], both in the x and y directions. 
In the x direction <math> \Sigma F = -kv_x = ma_x </math> and in the y direction <math> \Sigma F = -kv_y - mg = ma_y </math>.

This implies that:

<math> a_x = -\mu v_x = \frac{\mathrm{d}v_x}{\mathrm{d}t} </math> (1),

and

<math> a_y = -\mu v_y - g = \frac{\mathrm{d}v_y}{\mathrm{d}t} </math> (2)

Solving (1) is an elementary [[differential equation]], thus the steps leading to a unique solution for <var>v<sub>x</sub></var> and, subsequently, <var>x</var> will not be enumerated. Given the initial conditions <math> v_x = v_{x0} </math> (where <var>v<sub>x0</sub></var> is understood to be the x component of the initial velocity) and <math> x=0 </math> for <math> t=0 </math>:

<math> v_x = v_{x0} e^{-\mu t} </math> (1a)
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:<math> x(t) = \frac{v_{x0}}{\mu}\left(1-e^{-\mu t}\right) </math> (1b)
{{Collapse top|title=Derivation of vertical position}}
While (1) is solved much in the same way, (2) is of distinct interest because of its non-homogeneous nature. Hence, we will be extensively solving (2). Note that in this case the initial conditions are used <math> v_y=v_{y0} </math> and <math> y=0 </math> when <math> t=0 </math>.

<math> \frac{\mathrm{d}v_y}{\mathrm{d}t} = -\mu v_y - g </math> (2)

<math> \frac{\mathrm{d}v_y}{\mathrm{d}t} + \mu v_y = - g </math> (2a)

This first order, linear, non-homogeneous differential equation may be solved a number of ways; however, in this instance, it will be quicker to approach the solution via an [[integrating factor]] <math> e^{\int \mu \,\mathrm{d}t} </math>.

<math> e^{\mu t}(\frac{\mathrm{d}v_y}{\mathrm{d}t} + \mu v_y) = e^{\mu t}(-g) </math> (2c)

<math> (e^{\mu t}v_y)^\prime = e^{\mu t}(-g) </math> (2d)

<math> \int{(e^{\mu t}v_y)^\prime \,\mathrm{d}t} = e^{\mu t}v_y = \int{ e^{\mu t}(-g) \,\mathrm{d}t} </math> (2e)

<math> e^{\mu t}v_y = \frac{1}{\mu} e^{\mu t}(-g) + C </math>(2f)

<math> v_y = \frac{-g}{\mu} + Ce^{-\mu t} </math> (2g)

And by integration we find:

<math> y = -\frac{g}{\mu}t - \frac{1}{\mu}(v_{y0} + \frac{g}{\mu})e^{-\mu t} + C </math> (3)

Solving for our initial conditions:

<math> v_y(t) = -\frac{g}{\mu} + (v_{y0} + \frac{g}{\mu})e^{-\mu t} </math> (2h)

<math> y(t) = -\frac{g}{\mu}t - \frac{1}{\mu}(v_{y0} + \frac{g}{\mu})e^{-\mu t} + \frac{1}{\mu}(v_{y0} + \frac{g}{\mu}) </math> (3a)<br />
<br />
With a bit of algebra to simplify (3a):
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:<math> y(t) = -\frac{g}{\mu} t + \frac{1}{\mu} \left(v_{y0} + \frac{g}{\mu} \right) (1 - e^{-\mu t} ) </math> (3b)

{{Collapse top|title=Derivation of the time of flight}}
The total time of the journey in the presence of air resistance (more specifically, when <math>F_{air}=-kv</math>) can be calculated by the same strategy as above, namely, we solve the equation <math>y(t)=0</math>. While in the case of zero air resistance this equation can be solved elementarily, here we shall need the [[Lambert W function]]. The equation
<small><math>y(t)= -\frac{g}{\mu}t + \frac{1}{\mu}(v_{y0} + \frac{g}{\mu})(1 - e^{-\mu t}) = 0</math></small>
is of the form <small><math>c_1t+c_2+c_3e^{c_4t}=0</math></small>, and such an equation can be transformed into an equation solvable by the <small><math>W</math></small> function (see an example of such a transformation [[Lambert W function#Solving equations|here]]). Some algebra shows that the total time of flight, in closed form, is given as<ref name = 'Bernardo'>{{Cite journal | last1 = Reginald Cristian | first1 = Bernardo | last2 = Jose Perico | first2 = Esguerra | last3 = Jazmine Day | first3 = Vallejos | last4 = Jeff Jerard | first4 = Canda | title = Wind-influenced projectile motion | year = 2015 | volume = 36 | issue = 2 | journal = European Journal of Physics | page = 025016 | doi=10.1088/0143-0807/36/2/025016| bibcode = 2015EJPh...36b5016B | s2cid = 119601402 }}</ref>
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:<math>t=\frac{1}{\mu} \left(1+ \frac{\mu}{g} v_{y0} + W\bigl(-(1+ \frac{\mu}{g} v_{y0}) e^{-(1+ \frac{\mu}{g} v_{y0})} \bigr)\right)</math>.