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gibson:teaching:fall-2013:math445:lab11 [2013/12/05 08:35] szeto |
gibson:teaching:fall-2013:math445:lab11 [2013/12/05 09:10] (current) szeto |
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g = 9.81; % m/s^2, acceleration due to gravity | g = 9.81; % m/s^2, acceleration due to gravity | ||
. | . | ||
- | . | + | . |
- | . | + | . |
- | [t,x]= ode45(.... [ 0,0,v0*cos(theta), v0*sin(theta) ); | + | f = @(t,x) [ ....]; |
+ | |||
+ | [t,x]= ode45(f,[0:0.1:200] ,[ 0,0,v0*cos(theta), v0*sin(theta) ); | ||
+ | |||
+ | xdistance = interp1(x(50:end,2), x(50:end,1), 0); % This is how you can compute ''xdistance'' | ||
+ | % accurately from an ''x,y'' trajectory using interpolation. | ||
+ | % It will return the value of $x$ for which $y=0$, i.e. | ||
+ | % where the cannonball hits the ground. | ||
end | end | ||
</code> | </code> | ||
- | Do plotting and interpolating in another script or on the command line. | ||
- | You can compute ''xdistance'' accurately from an ''x,y'' trajectory using interpolation: | + | Once your function works, play around with the parameters ''v0, theta'' to answer the questions. |
+ | On the command line, you can USE this function like this. | ||
+ | <code> | ||
+ | xdistance = projectile(100,pi/2) | ||
+ | xdistance = projectile(100,pi/4) | ||
+ | xdistance = projectile(100,pi/8) | ||
+ | </code> | ||
- | xdistance = interpolate(x(50:end,2), x(50:end,1), 0); | + | If your xdistance is a NaN, you may have to increase the number 50 in the interp1 function (1000 might be good, that is roughly half the length of [0:0.1:200]). |
- | + | ||
- | That will return the value of $x$ for which $y=0$, i.e. where the cannonball hits the ground. Once your function works, play around with the parameters ''v0, theta'' to answer the questions. | + | |
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