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I would like to know how I could calculate the distance an object would fall with air resistance, assuming it's starting velocity is 0.

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    $\begingroup$ are you sure the tag "electrical resistance" is correct? Does your question have anything to do with electrical resistance that you've not mentioned? $\endgroup$ Sep 16 '16 at 15:49
  • $\begingroup$ @Sad_lab_rat Sorry! That was a mistake! $\endgroup$ Sep 16 '16 at 15:50
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    $\begingroup$ I think you should try to be more clear (and also show your efforts so far). If the object starts with $v=0$ it will just fall to the ground vertically... $\endgroup$
    – DelCrosB
    Sep 16 '16 at 15:52
  • $\begingroup$ @DelCrosB Yes, but I would like to know how far it would travel in, for example, 1 second including air resistance. I know how to calculate the distance of a free falling object without air resistance, d = 1/2g*t² and I do not know how I would include air resistance in there $\endgroup$ Sep 16 '16 at 15:55
  • $\begingroup$ @DelCrosB if I just dropped something down, and it would take 5 seconds to hit the ground, then it would have travelled about 122.63m, right? Now, I'm pretty sure that, because of air resistance, this would actually be a little lower. $\endgroup$ Sep 16 '16 at 16:00
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As mentioned in the comments, the resistive force is proportional to the velocity and let the constant of proportionality be $b$.

SO, $ma=mg-bv$

$a=g-\frac{bv}{m}$

Let $b/m=k$

Then $a=g-kv$

$dv/dt=g-kv$

This is a linear differential equation. Solving this for v in terms of t, we get. $ve^{kt}=\frac{ge^{kt}}{k} + c$

Applying the condition $v=0$ at $t=0$, we get $c=-g/k$

Thus, $$v=\frac{g(1-e^{-kt})}{k}$$

Write $v=ds/dt$

Take $ dt$ to the other side.

$$ds=\frac{g(1-e^{-kt})}{k}dt$$

Now, I leave this integration for you to solve. Solve this as indefinite integration and obtain the value of integration constant by putting $s=0$ for $ t=0$.

One question came to my mind while solving. Will the relation obtained between $s$ and $t$ follow the concept of terminal velocity?

Edit:

This relation is only approximate because in practical situations, $b$ does not remain constant for the whole journey. The value of $b$ can be approximated by conducting an experiment and using the relation between $s$ and $t$ . Suppose you release an object from a height $h$ (let) and it hits the ground in time $t$. Then, the value of $b$ can be easily determined.

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  • $\begingroup$ "Will the relation obtained between $s$ and $t$ follow the concept of terminal velocity?" Yes, just put $t=\infty$ into the $v(t)$ equation to get terminal velocity. $\endgroup$
    – Gert
    Sep 16 '16 at 17:36
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    $\begingroup$ Drag is proportional to the square of velocity. $\endgroup$ Sep 16 '16 at 17:43
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    $\begingroup$ @WhatRoughBeast: For small objects moving slowly (on the scale of dust motes or oil drops in the Millikan experiment), the effects of the air's viscosity are dominant, and so the drag force is linear. The OP is probably more interested in things like baseballs or ball bearings where the inertial drag effects are greater, but there are definitely regimes where the drag forces are linear. $\endgroup$ Sep 16 '16 at 18:03

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