How Does Air Resistance Modify the Equation for Falling Objects?

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SUMMARY

The discussion centers on the mathematical modeling of falling objects under the influence of air resistance, specifically using the equation y = y₀ - [t + (e^{-bt} - 1) / b] * g/b. Participants demonstrate that for short time intervals, this equation simplifies to y = y₀ - 0.5(g)(t²) through the application of the Taylor series expansion of e^{-bt}. The necessity of truncating the series at the t² term is emphasized, as higher-order terms become negligible for small values of t, thus maintaining the accuracy of the approximation.

PREREQUISITES
  • Understanding of basic calculus, specifically Taylor series expansion.
  • Familiarity with kinematic equations, particularly yf = yi(t) + 0.5(at²).
  • Knowledge of the effects of air resistance on falling objects.
  • Basic physics concepts related to gravity and free fall.
NEXT STEPS
  • Study the Taylor series expansion in detail, focusing on its applications in physics.
  • Explore the impact of air resistance on different shapes and sizes of falling objects.
  • Investigate numerical methods for solving differential equations related to motion with air resistance.
  • Learn about the limitations of using approximations in physical models.
USEFUL FOR

Physics students, educators, and anyone interested in the mathematical modeling of motion, particularly those studying the effects of air resistance on falling objects.

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

Show that for short times the eqn is reduced to

y=y_{o} - 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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