(************** Content-type: application/mathematica ************** CreatedBy='Mathematica 5.2' Mathematica-Compatible Notebook This notebook can be used with any Mathematica-compatible application, such as Mathematica, MathReader or Publicon. The data for the notebook starts with the line containing stars above. To get the notebook into a Mathematica-compatible application, do one of the following: * Save the data starting with the line of stars above into a file with a name ending in .nb, then open the file inside the application; * Copy the data starting with the line of stars above to the clipboard, then use the Paste menu command inside the application. Data for notebooks contains only printable 7-bit ASCII and can be sent directly in email or through ftp in text mode. Newlines can be CR, LF or CRLF (Unix, Macintosh or MS-DOS style). 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Vous lancez un projectile en l'air depuis l'origine avec une \ vitesse initiale ||", Cell[BoxData[ FormBox[ OverscriptBox[ FormBox[ SubscriptBox[ StyleBox["v", FontSlant->"Italic"], "0"], "TextForm"], FormBox["\[Rule]", "TextForm"]], TextForm]]], "||=", Cell[BoxData[ StyleBox[ SubscriptBox[ StyleBox["v", FontFamily->"Arial", FontSlant->"Italic"], "initiale"], FontFamily->"Arial"]]], " sous un angle \[Alpha]. \[CapitalEAcute]crivons \ l'acc\[EAcute]l\[EAcute]ration ", Cell[BoxData[ OverscriptBox[ FormBox[ StyleBox["g", FontSlant->"Italic"], "TextForm"], FormBox["\[Rule]", "TextForm"]]]], ", la vitesse initiale ", Cell[BoxData[ OverscriptBox[ FormBox[ SubscriptBox[ StyleBox["v", FontSlant->"Italic"], "0"], "TextForm"], FormBox["\[Rule]", "TextForm"]]]], " et les \[EAcute]quations param\[EAcute]triques de la trajectoire sous \ forme vectorielle :" }], "Text", FontFamily->"Arial"], Cell[BoxData[{ RowBox[{ RowBox[{ StyleBox[\(g = {0, gy}\), FontColor->RGBColor[0, 0, 1]], StyleBox[";", FontColor->RGBColor[0, 0, 1]], StyleBox[" ", FontColor->RGBColor[0, 0, 1]], StyleBox[\( (*\ vecteur\ acc\[EAcute]l\[EAcute]ration\ *) \), FontColor->RGBColor[0, 0, 1]], "\[IndentingNewLine]", StyleBox[\(v = {v0*Cos[alpha], v0*Sin[alpha] + g[\([2]\)]*t}\), FontColor->RGBColor[0, 0, 1]], StyleBox[";", FontColor->RGBColor[0, 0, 1]]}], StyleBox[" ", FontColor->RGBColor[0, 0, 1]], StyleBox[\( (*\ vecteur\ vitesse\ *) \), FontColor->RGBColor[0, 0, 1]]}], "\[IndentingNewLine]", RowBox[{ StyleBox[\(r = {v0*Cos[alpha] t, v0*Sin[alpha] t + 1/2 g[\([2]\)]*t\^2}\), FontColor->RGBColor[0, 0, 1]], StyleBox[";", FontColor->RGBColor[0, 0, 1]], StyleBox[\( (*\ r\ vecteur\ position\ *) \), FontColor->RGBColor[0, 0, 1]]}]}], "Input"], Cell["\<\ En \[EAcute]liminant le temps entre les deux \[EAcute]quations \ param\[EAcute]triques, on obtient l'\[EAcute]quation de la trajectoire \ :\ \>", "Text", FontFamily->"Arial"], Cell[CellGroupData[{ Cell[BoxData[ \(Solve[ Eliminate[{x == v0*Cos[alpha] t, y == v0*Sin[alpha] t + 1/2 g[\([2]\)]*t\^2}, t], y]\)], "Input"], Cell[BoxData[ \({{y \[Rule] \(Sec[alpha]\^2\ \((gy\ x\^2 + 2\ v0\^2\ x\ Cos[alpha]\ \ Sin[alpha])\)\)\/\(2\ v0\^2\)}}\)], "Output"] }, Open ]], Cell["Une autre syntaxe possible :", "Text", FormatType->TextForm, FontFamily->"Arial"], Cell[CellGroupData[{ Cell[BoxData[ \(Solve[{x == v0*Cos[alpha] t, y == v0*Sin[alpha] t + 1/2 g[\([2]\)]*t\^2}, y, t]\)], "Input"], Cell[BoxData[ \({{y \[Rule] \(x\ Sec[alpha]\^2\ \((gy\ x + v0\^2\ Sin[2\ \ alpha])\)\)\/\(2\ v0\^2\)}}\)], "Output"] }, Open ]], Cell[TextData[{ "Ces \[EAcute]quations permettent de r\[EAcute]soudre plusieurs probl\ \[EGrave]mes :\n1.\[NonBreakingSpace]Quelles sont les coordonn\[EAcute]es du \ sommet de la parabole ?\n2. Quelle est la port\[EAcute]e du tir ?\n3. Quel \ est, pour une vitesse initiale dont la grandeur est donn\[EAcute]e, l'angle \ de tir qui donne la plus grande port\[EAcute]e?\n4. Que vaut la \ port\[EAcute]e maximale ?\n5. Pour une vitesse initiale de grandeur fix\ \[EAcute]e, quelle valeur faut-il donner \[AGrave] \[Alpha] pour que le \ projectile atteigne un point ", StyleBox["P", FontSlant->"Italic"], " fix\[EAcute] d'avance dans le plan ", StyleBox["Oxy.\n", FontSlant->"Italic"], "6. Quelle est l'expression de l'enveloppe des tirs ?" }], "Text", FontFamily->"Arial"], Cell[TextData[{ "1. Au sommet ", StyleBox["S", FontSlant->"Italic"], " de la parabole, la composante ", Cell[BoxData[ \(TextForm\`v\_y\)], FontSlant->"Italic"], " de la vitesse est nulle. En posant ", Cell[BoxData[ FormBox[ StyleBox[\(v\_y\), FontSlant->"Italic"], TextForm]]], "=0, en r\[EAcute]solvant l'\[EAcute]quation donnant cette composante de la \ vitesse par rapport \[AGrave] ", StyleBox["t", FontSlant->"Italic"], " et en introduisant le r\[EAcute]sultat dans les \[EAcute]quations param\ \[EAcute]triques de la trajectoire, on obtient le temps cherch\[EAcute] et \ les coordonn\[EAcute]es du sommet :" }], "Text", FontFamily->"Arial"], Cell[CellGroupData[{ Cell[BoxData[{ \(Solve[v[\([2]\)] \[Equal] 0, t]\), "\[IndentingNewLine]", \(sommet = Flatten[r /. %]\)}], "Input"], Cell[BoxData[ \({{t \[Rule] \(-\(\(v0\ Sin[alpha]\)\/gy\)\)}}\)], "Output"], Cell[BoxData[ \({\(-\(\(v0\^2\ Cos[alpha]\ Sin[ alpha]\)\/gy\)\), \(-\(\(v0\^2\ Sin[alpha]\^2\)\/\(2\ \ gy\)\)\)}\)], "Output"] }, Open ]], Cell[TextData[{ "2. La port\[EAcute]e du tir (distance entre le point de d\[EAcute]part et \ le point o\[UGrave] le projectile touche le sol, suppos\[EAcute] horizontal) \ s'obtient en posant ", StyleBox["y", FontSlant->"Italic"], "=0 dans l'\[EAcute]quation param\[EAcute]trique donnant ", StyleBox["y", FontSlant->"Italic"], ", en r\[EAcute]solvant par rapport \[AGrave] ", StyleBox["t", FontSlant->"Italic"], " et en introduisant la solution retenue dans l'\[EAcute]quation donnant ", StyleBox["x", FontSlant->"Italic"], " :" }], "Text", FontFamily->"Arial"], Cell[CellGroupData[{ Cell[BoxData[{ \(\(Solve[r[\([2]\)] \[Equal] 0, t];\)\), "\[IndentingNewLine]", \(\(r[\([1]\)] /. %[\([2]\)];\)\), "\[IndentingNewLine]", \(portee = TrigReduce[%]\)}], "Input"], Cell[BoxData[ \(\(-\(\(v0\^2\ Sin[2\ alpha]\)\/gy\)\)\)], "Output"] }, Open ]], Cell[TextData[{ "3. La grandeur de la vitesse initiale \[EAcute]tant fix\[EAcute]e, la port\ \[EAcute]e est maximale lorsque ", StyleBox["sin", FontSlant->"Italic"], "(2\[Alpha]) prend sa plus grande valeur, c'est-\[AGrave]-dire lorsque 2\ \[Alpha]=90\[Degree], donc lorsque \[Alpha]=45\[Degree].\n4. La \ port\[EAcute]e maximale vaut ", Cell[BoxData[ FormBox[ SuperscriptBox[ FormBox[ SubscriptBox[ StyleBox["v", FontSlant->"Italic"], "0"], "TextForm"], "2"], TextForm]]], "/", Cell[BoxData[ \(TextForm\`g\_y\)], FontSlant->"Italic"], "." }], "Text", FontFamily->"Arial"], Cell[TextData[{ "5. Pour qu'un point ", StyleBox["P", FontSlant->"Italic"], " de coordonn\[EAcute]es (", Cell[BoxData[ FormBox[ StyleBox[\(p\_x\), FontSlant->"Italic"], TextForm]]], ", ", Cell[BoxData[ FormBox[ StyleBox[\(p\_y\), FontSlant->"Italic"], TextForm]]], ") soit atteint lorsque la grandeur ", Cell[BoxData[ FormBox[ SubscriptBox[ StyleBox["v", FontSlant->"Italic"], "0"], TextForm]]], " de la vitesse initiale est donn\[EAcute]e, il faut que ce point se trouve \ sur la trajectoire. Les coordonn\[EAcute]es du point doivent donc satisfaire \ l'\[EAcute]quation de la trajectoire. Pour obtenir une \[EAcute]quation du \ second degr\[EAcute] en ", StyleBox["tg", FontSlant->"Italic"], "(\[Alpha]) et la r\[EAcute]soudre par rapport \[AGrave] ", StyleBox["tg", FontSlant->"Italic"], "(\[Alpha]) nous op\[EAcute]rons la substitution. ", Cell[BoxData[ \(TextForm\`sec\^2\)]], "(\[Alpha])=1+", Cell[BoxData[ FormBox[ SuperscriptBox[ StyleBox["tg", FontSlant->"Italic"], "2"], TextForm]]], "(\[Alpha]). Nous obtenons alors deux solutions pour ", StyleBox["tg", FontSlant->"Italic"], "(\[Alpha]) :" }], "Text", FontFamily->"Arial"], Cell[CellGroupData[{ Cell[BoxData[{\(Solve[{xp == v0*Cos[alpha] t, yp == v0*Sin[alpha] t + 1/2 g[\([2]\)]*t\^2}, yp, t]\ ;\), "\[IndentingNewLine]", RowBox[{\(Simplify[%];\), " ", \( (*\ pour\ obtenir\ l' \[EAcute]quation\ de\ la\ trajectoire\ *) \)}], "\ \[IndentingNewLine]", \(% /. {Sec[alpha]^2 \[Rule] 1 + Tan[alpha]^2}\ ;\), "\[IndentingNewLine]", RowBox[{\(%[\([1, 1, 2]\)]\), "\[IndentingNewLine]", RowBox[{"(*", " ", RowBox[{ "substitution", " ", "pour", " ", "obtenir", " ", "une", " ", "\[EAcute]quation", " ", "du", " ", "second", " ", "degr\[EAcute]", " ", "en", " ", StyleBox["tg", FontSlant->"Italic"], \((\[Alpha])\)}], " ", "*)"}]}], "\[IndentingNewLine]", \(sol = Solve[yp \[Equal] %, Tan[alpha]]\)}], "Input"], Cell[BoxData[ \(xp\ Tan[ alpha] + \(gy\ xp\^2\ \((1 + Tan[alpha]\^2)\)\)\/\(2\ v0\^2\)\)], \ "Output"], Cell[BoxData[ \({{Tan[ alpha] \[Rule] \(\(-v0\^2\)\ xp - \@\(v0\^4\ xp\^2 - gy\^2\ xp\^4 \ + 2\ gy\ v0\^2\ xp\^2\ yp\)\)\/\(gy\ xp\^2\)}, {Tan[ alpha] \[Rule] \(\(-v0\^2\)\ xp + \@\(v0\^4\ xp\^2 - gy\^2\ xp\^4 \ + 2\ gy\ v0\^2\ xp\^2\ yp\)\)\/\(gy\ xp\^2\)}}\)], "Output"] }, Open ]], Cell[TextData[{ "Selon la valeur du discriminant, les deux solutions sont complexes, \ confondues ou diff\[EAcute]rentes, ce qui correspond respectivement \[AGrave] \ 0, 1 ou 2 solutions r\[EAcute]elles pour l'angle. Si le discriminant est n\ \[EAcute]gatif, le point ", StyleBox["P", FontSlant->"Italic"], " de coordonn\[EAcute]es (", Cell[BoxData[ FormBox[ StyleBox[\(p\_x\), FontSlant->"Italic"], TextForm]]], ", ", Cell[BoxData[ FormBox[ StyleBox[\(p\_y\), FontSlant->"Italic"], TextForm]]], ") est inaccessible; si le discriminant est nul, le point ", StyleBox["P", FontSlant->"Italic"], " est sur l'enveloppe et si le discriminant est positif, le point ", StyleBox["P", FontSlant->"Italic"], " peut \[EHat]tre atteint de deux mani\[EGrave]res. \n\n6. \ \[CapitalEAcute]galons le discriminant \[AGrave] 0, et r\[EAcute]solvons par \ rapport \[AGrave] (", StyleBox["x", FontSlant->"Italic"], ", ", StyleBox["y", FontSlant->"Italic"], "). Nous obtenons l'\[EAcute]quation de l'enveloppe des tajectoires:" }], "Text", FontFamily->"Arial"], Cell[CellGroupData[{ Cell[BoxData[{ \(Remove[enveloppe]\), "\[IndentingNewLine]", \(Off[Solve::svars]\), "\[IndentingNewLine]", \(\(Solve[ v0\^4\ xp\^2 - gy\^2\ xp\^4 + 2\ gy\ v0\^2\ xp\^2\ yp \[Equal] 0, \ {xp, yp}];\)\), "\[IndentingNewLine]", \(enveloppe[v0_] = %[\([3, 1, 2]\)]\)}], "Input"], Cell[BoxData[ \(\(\(-v0\^4\) + gy\^2\ xp\^2\)\/\(2\ gy\ v0\^2\)\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell["Examinons ce qu'on obtient avec des valeurs num\[EAcute]riques", \ "Subsection", FormatType->TextForm, FontFamily->"Arial"], Cell[BoxData[ \(\(vn = {gy \[Rule] \(-9.81\), v0 \[Rule] 30, alpha \[Rule] 30 Degree, xp \[Rule] 20, yp \[Rule] 10};\)\)], "Input"], Cell[CellGroupData[{ Cell["Sommet de la parabole", "Subsubsection", FormatType->TextForm, FontFamily->"Arial"], Cell[CellGroupData[{ Cell[BoxData[ \(sommet /. vn\)], "Input"], Cell[BoxData[ \({39.72593595341461`, 11.46788990825688`}\)], "Output"] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell["Port\[EAcute]e du tir", "Subsubsection", FormatType->TextForm, FontFamily->"Arial"], Cell[CellGroupData[{ Cell[BoxData[ \(portee /. vn\)], "Input"], Cell[BoxData[ \(79.45187190682923`\)], "Output"] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell["Angle de tir pour une vitesse initiale donn\[EAcute]e", "Subsubsection", FormatType->TextForm, FontFamily->"Arial"], Cell[CellGroupData[{ Cell[BoxData[{ \(\(ArcTan[tgalpha] /. vn;\)\), "\[IndentingNewLine]", \(%/Degree\)}], "Input"], Cell[BoxData[ \(33.260509997725066`\)], "Output"] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell["Enveloppe", "Subsubsection", FormatType->TextForm, FontFamily->"Arial"], Cell[CellGroupData[{ Cell[BoxData[ \(enveloppe[30] /. vn\)], "Input"], Cell[BoxData[ \(43.69155963302752`\)], "Output"] }, Open ]], Cell[BoxData[ \(\(Plot[enveloppe[30] /. vn, {xp, 0, 100}];\)\)], "Input"] }, Open ]] }, Open ]] }, Open ]] }, FrontEndVersion->"5.2 for Macintosh", ScreenRectangle->{{0, 994}, {0, 746}}, WindowSize->{887, 719}, WindowMargins->{{34, Automatic}, {-12, Automatic}}, ShowSelection->True, ShowCellLabel->False, Magnification->1.5, StyleDefinitions -> "Default.nb" ] (******************************************************************* Cached data follows. If you edit this Notebook file directly, not using Mathematica, you must remove the line containing CacheID at the top of the file. The cache data will then be recreated when you save this file from within Mathematica. *******************************************************************) (*CellTagsOutline CellTagsIndex->{} *) (*CellTagsIndex CellTagsIndex->{} *) (*NotebookFileOutline Notebook[{ Cell[CellGroupData[{ Cell[1776, 53, 51, 1, 115, "Title"], Cell[1830, 56, 1419, 50, 83, "Text"], Cell[3252, 108, 1044, 26, 68, "Input"], Cell[4299, 136, 186, 5, 39, "Text"], Cell[CellGroupData[{ Cell[4510, 145, 142, 3, 33, "Input"], Cell[4655, 150, 135, 2, 48, "Output"] }, Open ]], Cell[4805, 155, 91, 2, 39, "Text"], Cell[CellGroupData[{ Cell[4921, 161, 122, 2, 33, "Input"], Cell[5046, 165, 120, 2, 48, "Output"] }, Open ]], Cell[5181, 170, 790, 16, 190, "Text"], Cell[5974, 188, 708, 21, 82, "Text"], Cell[CellGroupData[{ Cell[6707, 213, 124, 2, 49, "Input"], Cell[6834, 217, 79, 1, 47, "Output"], Cell[6916, 220, 147, 3, 50, "Output"] }, Open ]], Cell[7078, 226, 597, 17, 82, "Text"], Cell[CellGroupData[{ Cell[7700, 247, 190, 3, 67, "Input"], Cell[7893, 252, 71, 1, 49, "Output"] }, Open ]], Cell[7979, 256, 680, 22, 82, "Text"], Cell[8662, 280, 1316, 43, 104, "Text"], Cell[CellGroupData[{ Cell[10003, 327, 852, 17, 120, "Input"], Cell[10858, 346, 116, 3, 48, "Output"], Cell[10977, 351, 300, 5, 94, "Output"] }, Open ]], Cell[11292, 359, 1140, 33, 168, "Text"], Cell[CellGroupData[{ Cell[12457, 396, 312, 6, 86, "Input"], Cell[12772, 404, 81, 1, 51, "Output"] }, Open ]], Cell[CellGroupData[{ Cell[12890, 410, 133, 3, 65, "Subsection"], Cell[13026, 415, 147, 2, 32, "Input"], Cell[CellGroupData[{ Cell[13198, 421, 93, 2, 52, "Subsubsection"], Cell[CellGroupData[{ Cell[13316, 427, 45, 1, 32, "Input"], Cell[13364, 430, 74, 1, 32, "Output"] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell[13487, 437, 93, 2, 52, "Subsubsection"], Cell[CellGroupData[{ Cell[13605, 443, 45, 1, 32, "Input"], Cell[13653, 446, 52, 1, 32, "Output"] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell[13754, 453, 125, 2, 52, "Subsubsection"], Cell[CellGroupData[{ Cell[13904, 459, 103, 2, 49, "Input"], Cell[14010, 463, 53, 1, 32, "Output"] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell[14112, 470, 81, 2, 52, "Subsubsection"], Cell[CellGroupData[{ Cell[14218, 476, 52, 1, 32, "Input"], Cell[14273, 479, 52, 1, 32, "Output"] }, Open ]], Cell[14340, 483, 77, 1, 32, "Input"] }, Open ]] }, Open ]] }, Open ]] } ] *) (******************************************************************* End of Mathematica Notebook file. *******************************************************************)