(************** 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). NOTE: If you modify the data for this notebook not in a Mathematica- compatible application, you must delete the line below containing the word CacheID, otherwise Mathematica-compatible applications may try to use invalid cache data. For more information on notebooks and Mathematica-compatible applications, contact Wolfram Research: web: http://www.wolfram.com email: info@wolfram.com phone: +1-217-398-0700 (U.S.) Notebook reader applications are available free of charge from Wolfram Research. *******************************************************************) (*CacheID: 232*) (*NotebookFileLineBreakTest NotebookFileLineBreakTest*) (*NotebookOptionsPosition[ 11793, 363]*) (*NotebookOutlinePosition[ 12513, 388]*) (* CellTagsIndexPosition[ 12469, 384]*) (*WindowFrame->Normal*) Notebook[{ Cell[CellGroupData[{ Cell[TextData[StyleBox["Projectile at an Angle \[EmptySet] Above Inclined \ Plane of Angle \[Theta]", FontSize->18]], "Title", TextAlignment->Center, FontSize->24, FontColor->RGBColor[0, 0, 1]], Cell[CellGroupData[{ Cell["Part A", "Subsection", FontSize->16, FontColor->RGBColor[0, 0, 1]], Cell[TextData[{ "The projectile is being fired at an angle \[EmptySet]+\[Theta] above the \ horizontal. So the trajectory equation is\n\ny(x) = \ tan(\[EmptySet]+\[Theta])x - g", Cell[BoxData[ FormBox[ SuperscriptBox[ StyleBox["x", FontWeight->"Plain", FontSlant->"Plain", FontTracking->"Plain", FontVariations->{"Underline"->False, "Outline"->False, "Shadow"->False, "StrikeThrough"->False, "Masked"->False, "CompatibilityType"->0, "RotationAngle"->0}], "2"], TraditionalForm]]], "/(2", Cell[BoxData[ FormBox[ StyleBox[\(v\_0\%2\), FontWeight->"Plain", FontSlant->"Plain", FontTracking->"Plain", FontVariations->{"Underline"->False, "Outline"->False, "Shadow"->False, "StrikeThrough"->False, "Masked"->False, "CompatibilityType"->0, "RotationAngle"->0}], TraditionalForm]]], Cell[BoxData[ \(TraditionalForm\`cos\^2\)]], "(\[EmptySet]+\[Theta])) where ", Cell[BoxData[ FormBox[ SubscriptBox[ StyleBox["v", FontWeight->"Plain", FontSlant->"Plain", FontTracking->"Plain", FontVariations->{"Underline"->False, "Outline"->False, "Shadow"->False, "StrikeThrough"->False, "Masked"->False, "CompatibilityType"->0, "RotationAngle"->0}], "0"], TraditionalForm]]], "is the speed.\n\nNow if the projectile hits the plane at coordinates \ (x,y), then we know the expression for tan\[Theta]\n\ntan\[Theta] = y/x \ which gives y = x tan\[Theta] . We can set these two expression for y equal \ to each other\n\nx tan\[Theta] = tan(\[EmptySet]+\[Theta])x - g", Cell[BoxData[ FormBox[ SuperscriptBox[ StyleBox["x", FontWeight->"Plain", FontSlant->"Plain", FontTracking->"Plain", FontVariations->{"Underline"->False, "Outline"->False, "Shadow"->False, "StrikeThrough"->False, "Masked"->False, "CompatibilityType"->0, "RotationAngle"->0}], "2"], TraditionalForm]]], "/(2", Cell[BoxData[ FormBox[ StyleBox[\(v\_0\%2\), FontWeight->"Plain", FontSlant->"Plain", FontTracking->"Plain", FontVariations->{"Underline"->False, "Outline"->False, "Shadow"->False, "StrikeThrough"->False, "Masked"->False, "CompatibilityType"->0, "RotationAngle"->0}], TraditionalForm]]], Cell[BoxData[ \(TraditionalForm\`cos\^2\)]], "(\[EmptySet]+\[Theta]))\n\nThis equation has a solution at x = 0, \ corresponding to y = 0, meaning when the projectile was fired.\n\nThere is a \ second solution when x is not 0, which we can derive by first dividing both \ sides by x\n\ntan\[Theta] = tan(\[EmptySet]+\[Theta]) - g", Cell[BoxData[ FormBox[ StyleBox["x", FontWeight->"Plain", FontSlant->"Plain", FontTracking->"Plain", FontVariations->{"Underline"->False, "Outline"->False, "Shadow"->False, "StrikeThrough"->False, "Masked"->False, "CompatibilityType"->0, "RotationAngle"->0}], TraditionalForm]]], "/(2", Cell[BoxData[ FormBox[ StyleBox[\(v\_0\%2\), FontWeight->"Plain", FontSlant->"Plain", FontTracking->"Plain", FontVariations->{"Underline"->False, "Outline"->False, "Shadow"->False, "StrikeThrough"->False, "Masked"->False, "CompatibilityType"->0, "RotationAngle"->0}], TraditionalForm]]], Cell[BoxData[ \(TraditionalForm\`cos\^2\)]], "(\[EmptySet]+\[Theta])) which leads to this equation for x on the \ left side\n\nx = [ tan(\[EmptySet]+\[Theta]) - tan\[Theta]]2", Cell[BoxData[ FormBox[ StyleBox[\(v\_0\%2\), FontWeight->"Plain", FontSlant->"Plain", FontTracking->"Plain", FontVariations->{"Underline"->False, "Outline"->False, "Shadow"->False, "StrikeThrough"->False, "Masked"->False, "CompatibilityType"->0, "RotationAngle"->0}], TraditionalForm]]], Cell[BoxData[ \(TraditionalForm\`cos\^2\)]], "(\[EmptySet]+\[Theta])\n\nLast, we can divide by cos\[Theta] to get the \ distance along the inclined plane (range)\n\nx/cos\[Theta] = [ \ tan(\[EmptySet]+\[Theta]) - tan\[Theta]]2", Cell[BoxData[ FormBox[ StyleBox[\(v\_0\%2\), FontWeight->"Plain", FontSlant->"Plain", FontTracking->"Plain", FontVariations->{"Underline"->False, "Outline"->False, "Shadow"->False, "StrikeThrough"->False, "Masked"->False, "CompatibilityType"->0, "RotationAngle"->0}], TraditionalForm]]], Cell[BoxData[ \(TraditionalForm\`cos\^2\)]], "(\[EmptySet]+\[Theta])/cos\[Theta]" }], "Text", FontSize->16, FontWeight->"Plain", FontColor->RGBColor[0, 0, 1], FontVariations->{"CompatibilityType"->0}] }, Open ]], Cell[CellGroupData[{ Cell["Part B", "Subsection", FontSize->16, FontColor->RGBColor[0, 0, 1]], Cell[TextData[{ "To get the maximum range, we need to take the derivative with respect to \ the angle \[EmptySet] of the right hand side of the last equation, and set \ that derivative to 0. There is a commercial computer program called ", StyleBox["Mathematica", FontSlant->"Italic"], " which can do derivatives and algebraic operations easily, although there \ is a learning curve of a few days time on getting the ", StyleBox["Mathematica", FontSlant->"Italic"], " syntax correct. We can simplify the problem by dropping the factor 2", Cell[BoxData[ FormBox[ StyleBox[\(v\_0\%2\), FontWeight->"Plain", FontSlant->"Plain", FontTracking->"Plain", FontVariations->{"Underline"->False, "Outline"->False, "Shadow"->False, "StrikeThrough"->False, "Masked"->False, "CompatibilityType"->0, "RotationAngle"->0}], TraditionalForm]]], "/cos\[Theta] which do not depend upon the angle \[EmptySet]. So we are \ left with the task of taking the derivative of the expression \ [tan(\[EmptySet]+\[Theta]) - tan\[Theta]]", Cell[BoxData[ \(TraditionalForm\`cos\^2\)]], "(\[EmptySet]+\[Theta]) . In the following, the text in", StyleBox[" RED ", FontWeight->"Bold", FontColor->RGBColor[1, 0, 0]], StyleBox["is an input command to Mathematica, while the text in ", FontVariations->{"CompatibilityType"->0}], StyleBox["MAGENTA ", FontWeight->"Bold", FontColor->RGBColor[1, 0, 1], FontVariations->{"CompatibilityType"->0}], StyleBox["is the output result of that command. First we find the \ derivative with respect to ", FontVariations->{"CompatibilityType"->0}], "the angle \[EmptySet]", StyleBox[", and the ", FontVariations->{"CompatibilityType"->0}], StyleBox["Mathematica", FontSlant->"Italic", FontVariations->{"CompatibilityType"->0}], StyleBox[" command is", FontVariations->{"CompatibilityType"->0}] }], "Text", FontSize->16, FontColor->RGBColor[0, 0, 1]], Cell[CellGroupData[{ Cell[BoxData[ StyleBox[\(D[\((Tan[\[EmptySet] + \[Theta]]\ - \ Tan[\[Theta]])\)*\((Cos[\[EmptySet] + \[Theta]])\)\^2, \ \[EmptySet]]\), FontColor->RGBColor[1, 0, 0]]], "Input", FontSize->16, FontColor->RGBColor[0, 0, 1]], Cell[BoxData[ \(1 - 2\ Cos[\[Theta] + \[EmptySet]]\ Sin[\[Theta] + \[EmptySet]]\ \((\(-Tan[\ \[Theta]]\) + Tan[\[Theta] + \[EmptySet]])\)\)], "Output", FontSize->16, FontColor->RGBColor[1, 0, 1]] }, Open ]], Cell["\<\ The above is the derivative result, and we can simplify it using \ trig identities to obtain\ \>", "Text", FontSize->16, FontColor->RGBColor[0, 0, 1]], Cell[CellGroupData[{ Cell[BoxData[ StyleBox[\(FullSimplify[ 1 - 2\ Cos[\[Theta] + \[EmptySet]]\ Sin[\[Theta] + \[EmptySet]]\ \ \((\(-Tan[\[Theta]]\) + Tan[\[Theta] + \[EmptySet]])\)]\), FontColor->RGBColor[1, 0, 0]]], "Input", FontSize->16, FontColor->RGBColor[0, 0, 1]], Cell[BoxData[ \(Cos[\[Theta] + 2\ \[EmptySet]]\ Sec[\[Theta]]\)], "Output", FontSize->16, FontColor->RGBColor[1, 0, 1]] }, Open ]], Cell["\<\ Lastly, we can set the above result equal to 0 in order to find the \ value of \[EmptySet] which gives a maximum. By simple inspection you can see \ that \[Theta]+2\[EmptySet]=\[Pi]/2 is the correct answer since it makes the \ cosine term to be 0.\ \>", "Text", FontSize->16, FontColor->RGBColor[0, 0, 1]], Cell[CellGroupData[{ Cell[BoxData[ StyleBox[\(Solve[ Cos[\[Theta] + 2\ \[EmptySet]]\ Sec[\[Theta]] \[Equal] 0, \ \[EmptySet]]\), FontColor->RGBColor[1, 0, 0]]], "Input", FontSize->16, FontColor->RGBColor[0, 0, 1]], Cell[BoxData[ \({{\[EmptySet] \[Rule] 1\/2\ \((\(-\(\[Pi]\/2\)\) - \[Theta])\)}, {\[EmptySet] \[Rule] 1\/2\ \((\[Pi]\/2 - \[Theta])\)}}\)], "Output", FontSize->16, FontColor->RGBColor[1, 0, 1]] }, Open ]], Cell["\<\ The second (positive) solution \[EmptySet] = \[Pi]/4 - \[Theta]/2 \ is what we are looking for, since negative angles would not have any physical \ meaning in this problem.\ \>", "Text", FontSize->16, FontColor->RGBColor[0, 0, 1]] }, Open ]] }, Open ]] }, FrontEndVersion->"5.2 for Macintosh", ScreenRectangle->{{0, 1241}, {0, 1002}}, WindowToolbars->"EditBar", WindowSize->{791, 738}, WindowMargins->{{201, Automatic}, {Automatic, 42}}, PrintingCopies->1, PrintingPageRange->{1, Automatic} ] (******************************************************************* 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, 201, 5, 73, "Title"], Cell[CellGroupData[{ Cell[2002, 62, 76, 2, 38, "Subsection"], Cell[2081, 66, 5347, 156, 434, "Text"] }, Open ]], Cell[CellGroupData[{ Cell[7465, 227, 76, 2, 38, "Subsection"], Cell[7544, 231, 2058, 52, 174, "Text"], Cell[CellGroupData[{ Cell[9627, 287, 250, 6, 32, "Input"], Cell[9880, 295, 210, 5, 30, "Output"] }, Open ]], Cell[10105, 303, 164, 5, 34, "Text"], Cell[CellGroupData[{ Cell[10294, 312, 271, 6, 30, "Input"], Cell[10568, 320, 127, 3, 30, "Output"] }, Open ]], Cell[10710, 326, 319, 7, 54, "Text"], Cell[CellGroupData[{ Cell[11054, 337, 225, 6, 30, "Input"], Cell[11282, 345, 223, 5, 47, "Output"] }, Open ]], Cell[11520, 353, 245, 6, 54, "Text"] }, Open ]] }, Open ]] } ] *) (******************************************************************* End of Mathematica Notebook file. *******************************************************************)