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I´m working on a free fall problem and I got an equation with variable $k$ when the situation has air resistance. Then I compared with no air resistance equation ($-4.9$, considering t variable). So, I got this:

113.68t/k + (1345.6/(k^2))*(1 - Exp^(-kt/11.6)) == -4.9t^2

I tried NSolve, DSolve, NDSolve to find the solutions of $k$, but none of them could answer it.

Until now i got this

 y := -4.9 t^2
a = Plot[113.68t/k-(1345.6/(k^2))*(1-e^(-k*t/11.6))= -4.9 t^2, {t, 0, 1.4}, 
  PlotStyle -> Green]
teorico = Plot[y, {t, 0, 1.4}];
Show[teorico, a, PlotRange -> All]
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  • 2
    $\begingroup$ You have several syntactic errors, such as Exp^(...) instead of Exp[...], -4,9 instead of -4.9. But your fundamental problem is that your equation must be incorrect, since there is no solution. $\endgroup$ – David G. Stork Apr 26 '16 at 18:01
  • $\begingroup$ Note that DSolve and NDSolve are for solving differential equations (either analytically or numerically.) One wouldn't expect them to be of any help here. $\endgroup$ – Michael Seifert Apr 26 '16 at 18:06
  • $\begingroup$ That's a different equation; there's probably a solution to this one. (Your original one had a minus sign on the second term.) However, your equation still has the syntax problems pointed out by @DavidG.Stork concerning the exponential function. $\endgroup$ – Michael Seifert Apr 26 '16 at 18:29
  • $\begingroup$ Put your cursor on Exp and push F1 . You managed to invent two different incorrect expression syntaxes for the exponential. $\endgroup$ – george2079 Apr 26 '16 at 18:38
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There is no solution to your equation. If you specify Reals for k Mathematica outputs an empty list as solution:

NSolve[113.68/k + 1345.6/k^2*(1 - Exp[-k/11.6]) == -4.9, k, Reals]
(*{}*)

If you plot the righ and left hand side of the equation they don't cut each other:

Plot[{113.68/k + 1345.6/k^2*(1 - Exp[-k/11.6]), -4.9}, {k, -60, 60}]

enter image description here

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The corrected equation can be solved using FindRoot:

FindRoot[(113.68 t/k - (1345.6/(k^2))*(1 - Exp[-k*t/11.6]) == -4.9 t^2) /. t -> 1, {k, 0.5}]
(* {k -> 0.235142} *)

This solves the equation for the value of $k$ corresponding to $t$ = 1 second. FindRoot requires an initial "guess" for the value of $k$; in this case, I've used $k = 0.5$.

Other choices of $t$ will yield different values of $k$. In fact, your equation reduces to $kt = \alpha$, where $\alpha$ is the root of the equation $$ 113.68/\alpha - (1345.6/(\alpha^2))*(1 - e^{-\alpha/11.6}) = -4.9 $$ which, as noted above, yields $\alpha \approx 0.235142$. For a given value of $t$, $k$ will then be equal to $\alpha/t$.


For the original equation (with a minus sign on the second term), no solution exists. We note from the form of the equation that the left-hand side will be strictly positive for positive values of $k$; so if the left-hand side is equal to -4.9, then we must have $k <0$. But if we run

FindMaximum[{113.68/k + (1345.6/(k^2))*(1 - Exp[-k/11.6]), k < 0}, k]

we obtain

(* {-20.8733, {k -> -22.9611}} *)

and so the function never attains a value greater than -20.8733 for any negative value of $k$.

If you don't believe me, here's the graph:

Plot[113.68/k + (1345.6/(k^2))*(1 - Exp[-k/11.6]), {k, -100, 100}]

enter image description here

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  • $\begingroup$ Note the correct form of the exponential function in my code, as was pointed out in the comments. $\endgroup$ – Michael Seifert Apr 26 '16 at 18:02

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