{"id":4581,"date":"2017-12-07T11:35:09","date_gmt":"2017-12-07T16:35:09","guid":{"rendered":"http:\/\/blog.espol.edu.ec\/matg1013\/?p=4581"},"modified":"2026-04-05T20:08:58","modified_gmt":"2026-04-06T01:08:58","slug":"s2eva2007tii_t2_an-lanzamiento-vertical-proyectil","status":"publish","type":"post","link":"https:\/\/blog.espol.edu.ec\/algoritmos101\/mn-s2eva10\/s2eva2007tii_t2_an-lanzamiento-vertical-proyectil\/","title":{"rendered":"s2Eva2007TII_T2_AN EDO Lanzamiento vertical proyectil"},"content":{"rendered":"\n

Ejercicio<\/strong>: 2Eva2007TII_T2_AN EDO Lanzamiento vertical proyectil<\/a><\/p>\n\n\n\n

la ecuaci\u00f3n del problema:<\/p>\n\n\n m \\frac{\\delta v}{\\delta t} = -mg - kv|v|<\/span>\n\n\n\n

se despeja:<\/p>\n\n\n\\frac{\\delta v}{\\delta t} = -g - \\frac{k}{m}v|v|<\/span>\n\n\n\n

y usando los valores indicados en el enunciado:<\/p>\n\n\n\\frac{\\delta v}{\\delta t} = -9,8 - \\frac{0.002}{0.11}v|v|<\/span>\n\n\n\n

con valores iniciales de:<\/p>\n\n\n\n

t0<\/sub> = 0 , v0<\/sub> = 8 , h=0.2<\/p>\n\n\n\n

Como muestra inicial, se usa Runge-Kutta de 2do Orden<\/strong><\/em><\/p>\n\n\n\n

iteraci\u00f3n 1<\/h4>\n\n\nK_1 = h\\frac{\\delta v}{\\delta t}(0, 8)<\/span>\n\n\n= 0.2[-9,8 - \\frac{0.002}{0.11}8|8|] = -2.1927 <\/span>\n\n\nK_2 = h\\frac{\\delta v}{\\delta t}(0+0.2, 8 -2.1927 ) <\/span>\n\n\n= 0.2[-9,8 - \\frac{0.002}{0.11}(8 -2.1927)|8 -2.1927|] =-2.0826 <\/span>\n\n\nv_1 = -9,8 +\\frac{-2.1927-2.0826 }{2} = 5.8623<\/span>\n\n\nt_1 = t_0 + h = 0 + 0.2 = 0.2<\/span>\n\n\nerror = O(h^3) = O(0.2^3) = O(0.008)<\/span>\n\n\n\n

iteraci\u00f3n 2<\/h4>\n\n\nK_1 = h\\frac{\\delta v}{\\delta t}(0.2, 5.8623)<\/span>\n\n\n= 0.2[-9,8 - \\frac{0.002}{0.11}(5.8623)|5.8623|] = -2.085 <\/span>\n\n\nK_2 = h\\frac{\\delta v}{\\delta t}(0+0.2, 5.8623 -2.085) <\/span>\n\n\n= 0.2[-9,8 - \\frac{0.002}{0.11}(5.8623 -2.085)|5.8623 -2.085|] =-2.0119<\/span>\n\n\nv_2 = -9,8 +\\frac{-2.085-2.0119}{2} = 3.8139<\/span>\n\n\nt_2 = t_1 + h = 0.2 + 0.2 = 0.4<\/span>\n\n\n\n

iteraci\u00f3n 3<\/h4>\n\n\nK_1 = h\\frac{\\delta v}{\\delta t}(0.4, 3.8139)<\/span>\n\n\n= 0.2[-9,8 - \\frac{0.002}{0.11}( 3.8139)| 3.8139|] = -2.0129 <\/span>\n\n\nK_2 = h\\frac{\\delta v}{\\delta t}(0+0.2, 3.8139 -2.0129) <\/span>\n\n\n= 0.2[-9,8 - \\frac{0.002}{0.11}(3.8139 -2.0129)|3.8139 -2.0129|] =-1.9718<\/span>\n\n\nv_3 = -9,8 +\\frac{-2.0129-1.9718}{2} = 1.8215<\/span>\n\n\nt_3 = t_2 + h = 0.4 + 0.2 = 0.6<\/span>\n\n\n\n
\n\n\n\n

Tabla y gr\u00e1fica del ejercicio para todo el intervalo:<\/p>\n\n\n\n

 [xi,      yi,     K1,     K2    ]\n[[ 0.      8.      0.      0.    ]\n [ 0.2     5.8623 -2.1927 -2.0826]\n [ 0.4     3.8139 -2.085  -2.0119]\n [ 0.6     1.8215 -2.0129 -1.9718]\n [ 0.8    -0.1444 -1.9721 -1.9599]\n [ 1.     -2.0964 -1.9599 -1.9439]]<\/code><\/pre>\n\n\n\n
\"2EIIT2007T2<\/figure>\n\n\n\n
Algoritmo en Python<\/h2>\n\n\n
\n# 3Eva_IT2009_T2 EDO Taylor Seno(x)\nimport numpy as np\nimport matplotlib.pyplot as plt\n\n# INGRESO\nd1y = lambda t,v: -9.8-(0.002\/0.11)*v*np.abs(v)\nx0 = 0\ny0 = 8\nh = 0.2\na = 0\nb = 1\n\n# PROCEDIMIENTO\nmuestras = int((b -a)\/h)+1\ntabla = np.zeros(shape=(muestras,4),dtype=float)\n\ni = 0\nxi = x0\nyi = y0\ntabla[i,:] = [xi,yi,0,0]\n\ni = i+1\nwhile not(i>=muestras):\n    K1 = h*d1y(xi,yi)\n    K2 = h*d1y(xi+h,yi+K1)\n    yi = yi + (K1+K2)\/2\n    xi = xi +h\n    tabla[i,:] = [xi,yi,K1,K2]\n    i = i+1\n# vector para gr\u00e1fica\nxg = tabla[:,0]\nyg = tabla[:,1]\n\n# SALIDA\n# muestra 4 decimales\nnp.set_printoptions(precision=4)\nprint(' [xi, yi, K1, K2]')\nprint(tabla)\n# Gr\u00e1fica\nplt.plot(xg,yg)\nplt.xlabel('ti')\nplt.ylabel('yi')\nplt.grid()\nplt.show()\n<\/pre><\/div>\n\n\n
Tarea<\/strong>: Realizar iteraciones para Runge-Kutta de 4to Orden<\/p>\n","protected":false},"excerpt":{"rendered":"
Ejercicio: 2Eva2007TII_T2_AN EDO Lanzamiento vertical proyectil la ecuaci\u00f3n del problema: se despeja: y usando los valores indicados en el enunciado: con valores iniciales de: t0 = 0 , v0 = 8 , h=0.2 Como muestra inicial, se usa Runge-Kutta de 2do Orden iteraci\u00f3n 1 iteraci\u00f3n 2 iteraci\u00f3n 3 Tabla y gr\u00e1fica del ejercicio para todo […]<\/p>\n","protected":false},"author":8043,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"wp-custom-template-entrada-mn-ejemplo","format":"standard","meta":{"footnotes":""},"categories":[47],"tags":[58,54],"class_list":["post-4581","post","type-post","status-publish","format-standard","hentry","category-mn-s2eva10","tag-ejemplos-python","tag-mnumericos"],"_links":{"self":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/4581","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/users\/8043"}],"replies":[{"embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/comments?post=4581"}],"version-history":[{"count":3,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/4581\/revisions"}],"predecessor-version":[{"id":23850,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/4581\/revisions\/23850"}],"wp:attachment":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/media?parent=4581"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/categories?post=4581"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/tags?post=4581"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}