Mechanics of practical situations

So, as we all know, in an infinite medium, the drag (viscous) is given by

$6\pi \eta rv$

now, this is directed exactly opposite to velocity, clearly, this being linearly proportional to velocity, we can break into components and write

${ F }_{ drag-x }=6\pi \eta r{ v }_{ x }\\ \\ { F }_{ drag-y }=6\pi \eta r{ v }_{ y }$

Now let us not forget gravity as well , and write the equations of motion as

$\\ -m\ddot { x } =\quad 6\pi \eta r\dot { x } \\ \\ -m\ddot { y } =\quad 6\pi \eta r\dot { y } +mg$

Integrate the y- part twice (dont forget to put proper boundary conditions)

${ v }_{ y }(0)=10, { v }_{ x }(0)=10, x(0)=0, y(0)=0$

and get

$\frac { (k{ v }_{ y }+g)(1-{ e }^{ -tk }) }{ { k }^{ 2 } } -\frac { gt }{ k }$

where $k=\frac { 6\pi \eta r }{ m }$

(if you dont believe me, you may use taylor series expansion for exponential function which is ${ e }^{ x }=1+\frac { x }{ 1 } +\frac { { x }^{ 2 } }{ 2 } +.....$ neglecting higher order terms and you shall see that it reduces to :

${ v }_{ y }t-\frac { g }{ 2 } { t }^{ 2 }$

Now equate $y$ to $0$ , (it comes back to ground)

use $g=10$ and input values,

and after thanking Ronak for giving really nice values you get the expression

$t=11(1-{ e }^{ -\frac { t }{ 10 } })$

now using a graph plotter to get ( the graph is what i have uploaded, the point is marked)

$t= 1.937$ and also more usably $11(1-e^{\frac{-t}{10}})=1.937$

Now integrate the $x$ equation twice to get

$x=\frac { { v }_{ x } }{ k } (1-{ e }^{ -\frac { t }{ 10 } })=\quad \frac { 100 }{ 11 } (1.937)$

which is $17.6090 m$

Now the simple model (no air , nothing, all ideal, way) gives $20m$

Calculating percent gives us a deviation of $11.95$ percent

add $0.5$ and box it to get $12$

(please notify of any printing errors) (or correct them if you are moderator)