How Does Air Resistance Modify the Equation for Falling Objects?

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Homework Help Overview

The discussion revolves around the effect of air resistance on the motion of falling objects, specifically examining how it modifies the equation for height over time. The original poster presents a formula that incorporates air resistance and seeks to demonstrate its reduction to a simpler form for short time intervals.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the use of Taylor series expansion to simplify the exponential term in the equation. There are attempts to relate the derived equation to known kinematic equations. Questions arise about the necessity of the Taylor expansion and the significance of truncating the series at the second-order term.

Discussion Status

Some participants have provided guidance on using the Taylor expansion and its implications for small time intervals. The original poster has successfully derived the reduced equation but seeks further clarification on the mathematical reasoning behind the expansion and the choice of terms included.

Contextual Notes

There is an emphasis on understanding the behavior of the equation as time approaches zero, with discussions about the accuracy of approximations and the relevance of higher-order terms in the context of the problem.

affans
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Homework Statement



The effect of air resistance is to slow down an object. It can be shown that the height of a falling object is given by the following:

y=[tex]y_{o}[/tex] - [t + ([tex]e^{-bt}[/tex] - 1) / b] * g/b.

Show that for short times the eqn is reduced to

y=[tex]y_{o}[/tex] - 0.5(g)(t^2)

Homework Equations



I think the regular distance formula yf = yi(t) + 0.5(at^2) is what i need here because the reduced eqn resembles it very cloesely.


The Attempt at a Solution


I've tried to do a lim as t approaces 0 on the first eqn. I've tried to equate the second eqn with the first eqn. I've been at it for a couple of hours now.

ANY help would be appreciated.
 
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Try the Taylor expansion of exp(-bt) and ignore terms past t2
 
The distance formula yf = yi(t) + 0.5(at^2) is actually the same as y=y0 - 0.5(g)(t^2), the equation the question wants you to prove, since a=-g.

Do you know what the Taylor series expansion of e^x is? If you do, expand e^-bt, discard higher-ordered terms, and y will reduce to y0 - 0.5(g)(t^2).
 
Oops, I posted my answer before I saw rock.freak's.
 
Hi,
thankyou very much. I have gotten the reduced equation.

but can someone explain to me why the taylor expansion was needed? what does the expansion have to do with "time being very short"?
also, why are only the terms until t^2 needed?

thanks
 
affans said:
Hi,
thankyou very much. I have gotten the reduced equation.

but can someone explain to me why the taylor expansion was needed? what does the expansion have to do with "time being very short"?
also, why are only the terms until t^2 needed?

thanks

if t is small, terms like t3,t4 and higher will give even smaller numbers. So depending on the degree of accuracy, these numbers don't affect the desired accuracy.
 
so if i have understoon taylor series correctly, it just means the

function e^x (or in my case e^-bt) can be REWRITTEN as a sum of individual terms given by the taylors series. Am i correct?

and if I am correct then in my question, the higher the degree on t, the smaller the number.

so my third question is why stop at t^2?
 
"function e^x (or in my case e^-bt) can be REWRITTEN as a sum of individual terms given by the taylors series. Am i correct?"

Yup. The more terms you include, the more accurate the approximation. The Taylor series expansion becomes infinitely accurate with an infinite number of terms.

"and if I am correct then in my question, the higher the degree on t, the smaller the number."

Exactly.

"so my third question is why stop at t^2?"

Because that's the level of approximation that gives you y=y0 - 0.5(g)(t^2). If you include more terms, you'll get a more accurate equation, but it won't be the same as the free-fall equation.
 

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