6_pdfsam_1 - % % Initial values g = 32.2; % gravity, ft/s^2...

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3. Computational Method: Using the model developed above, expressions for the desired results can be obtained. The object will hit the ground when its vertical position is zero y ( t )= vt sin θ 1 2 gt 2 =0 which can be solved to yield two values of time t =0 , 2 v sin θ g The second of the two solutions indicates that the object will return to the ground at the time t g = 2 v sin θ g The horizontal position (distance of travel) at this time is x ( t g )= vt g cos θ 4. Computational Implementation: The equations deFned in the computational method can be readily implemented using Mat- lab . The commands used in the following solution will be discussed in detail later in the course, but observe that the Matlab steps match closely to the solution steps from the computational method. % Flight trajectory computation
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Unformatted text preview: % % Initial values g = 32.2; % gravity, ft/s^2 v = 50 * 5280/3600; % launch velocity, ft/s theta = 30 * pi/180; % launch angle, radians % Compute and display results disp(’time of flight (s):’) % label for time of flight tg = 2 * v * sin(theta)/g % time to return to ground, s disp(’distance traveled (ft):’) % label for distance xg = v * cos(theta) * tg % distance traveled % Compute and plot flight trajectory t = linspace(0,tg,256); x = v * cos(theta) * t; y = v * sin(theta) * t - g/2 * t.^2; plot(x,y), axis equal, axis([ 0 150 0 30 ]), grid, . .. xlabel(’Distance (ft)’), ylabel(’Height (ft)’), title(’Flight Trajectory’) Comments: 6...
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This note was uploaded on 03/07/2010 for the course ENG 101 taught by Professor Chang during the Summer '09 term at 東京国際大学.

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