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I have a equation of motion v(t), that is integrated (t=0) to x(t). This symbolizes a trajectory of a thrown object. I want to create compared Plots for x(t) with different values for beta. When I try to Plot this, I always get a straigt line instead a parabola. Could someone help ?

Friction force (Stokes): F = -beta*v

Alternatively how could x(t) could be calculated with DSolve from v(t) ?

thanks so much!

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I tried this:

m = 1;
g = -9.81;
eZ = 1;
beta = 4;
v0 = 18;


x[t_] := (m*v0)/beta*Cos[alpha]*(1 - e^(-(beta/m)*t));
Plot[x, {x, -10, 10}]
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closed as off-topic by Szabolcs, Edmund, Henrik Schumacher, m_goldberg, Daniel Lichtblau Jan 5 at 16:24

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  • 2
    $\begingroup$ You need to specific about what doesn't work, otherwise the question will be closed as incomplete. What did you do, what did you expect to happen, and what happened instead? $\endgroup$ – Szabolcs Jan 5 at 11:12
  • $\begingroup$ x is a function of t. You cannot plot x for x. You must plot x[t] for t. $\endgroup$ – Edmund Jan 5 at 11:34
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    $\begingroup$ Probably you mean the Euler number by e. Within Mathematica, this number is called E (hint: all built-in symbols start with a capital letter). $\endgroup$ – Henrik Schumacher Jan 5 at 11:39
  • $\begingroup$ eZ should be the unit vector for z $\endgroup$ – Tom Jan 5 at 11:43
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Try this

Clear[beta]
m = 1;
g = -9.81;
alpha = 2;
bmin = 1;
bmax = 10;
delb = 1;
v0 = 18;
z0 = 0;


x[t_] := (m*v0)/beta*Cos[alpha]*(1 - E^(-(beta/m)*t));
z[t_] := z0 - m g/beta + (m v0/beta Sin[alpha] + m^2 g/beta^2) (1 - 
 E^(-beta/m t))
gr1 = Table[Plot[x[t], {t, -1, 1}], {beta, bmin, bmax, delb}];
Show[gr1]
gr2 = Table[ParametricPlot[{x[t], z[t]}, {t, -1, 1}], {beta, bmin, bmax, delb}];
Show[gr2]
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  • $\begingroup$ looks perfect so far, thank you! For variation of beta and different plots, I have to remove the underlines and doublepoints in the equotations ? $\endgroup$ – Tom Jan 5 at 13:11
  • $\begingroup$ No. Those are essential elements for the function definition. $\endgroup$ – Cesareo Jan 5 at 13:26
  • $\begingroup$ Now, I have it, thank you :) $\endgroup$ – Tom Jan 5 at 13:30
  • $\begingroup$ How could I show different values of beta in one plot ? $\endgroup$ – Tom Jan 6 at 15:14
  • $\begingroup$ @Tom I introduced some modifications to show a diversity in $\beta$ $\endgroup$ – Cesareo Jan 6 at 18:02

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