gibson:teaching:fall-2013:math445:lab11
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gibson:teaching:fall-2013:math445:lab11 [2013/12/05 08:33] szeto |
gibson:teaching:fall-2013:math445:lab11 [2013/12/05 09:10] (current) szeto |
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| Hints: It's probably easiest to answer (a)-%%(c)%% if you turn your script into a function ''xdistance = projectile(v0, theta)'' that returns the distance the projectile travels as a function of the initial velocity and angle. | Hints: It's probably easiest to answer (a)-%%(c)%% if you turn your script into a function ''xdistance = projectile(v0, theta)'' that returns the distance the projectile travels as a function of the initial velocity and angle. | ||
| In other words, create a new function "projectile.m" | In other words, create a new function "projectile.m" | ||
| - | %%%%%%%%%%%% | + | <code> |
| function xdistance = projectile(v0,theta) | function xdistance = projectile(v0,theta) | ||
| Line 45: | Line 45: | ||
| 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> |
| - | Do plotting and interpolating in another script or on the command line. | + | |
| + | 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> | ||
| - | You can compute ''xdistance'' accurately from an ''x,y'' trajectory using interpolation: | ||
| - | 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. | + | |
| | | ||
gibson/teaching/fall-2013/math445/lab11.1386261210.txt.gz · Last modified: 2013/12/05 08:33 by szeto
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