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: <P>You don't need to worry about unit vectors as it has been complicated by the Hawking forum fool who can't even decide what a uniform g field means. I'll demonstrate how to get the range-angle equation and you can work it from there. You have</P><P>x = (v<SUB>0</SUB>cos<FONT FACE="Symbol">q</FONT>)t</P><P>y = (v<SUB>0</SUB>sin<FONT FACE="Symbol">q</FONT>)t - (1/2)gt<SUP>2</P></SUP><P>It lands at y = 0 and so you have</P><P>0 = (v<SUB>0</SUB>sin<FONT FACE="Symbol">q</FONT>)t - (1/2)gt<SUP>2</P></SUP><P>0 = t[v<SUB>0</SUB>sin<FONT FACE="Symbol">q</FONT> - (1/2)gt]</P><P>If it started at t = 0 then the other solution is</P><P>t = (2/g)v<SUB>0</SUB>sin<FONT FACE="Symbol">q</FONT> </P><P>Using this in the first equation at the top of the page results in</P><P>x = (v<SUB>0</SUB>cos<FONT FACE="Symbol">q</FONT>)(2/g)v<SUB>0</SUB>sin<FONT FACE="Symbol">q</FONT> </P><P>which with a trig identity simplifies to</P><P ALIGN="CENTER">x = (v<SUB>0</SUB><SUP>2</SUP>/g)sin(2<FONT FACE="Symbol">q</FONT>)</P><P>This equation will result in two angles for a given range and velocity except for the particular combination that gives you <FONT FACE="Symbol">q</FONT> = <FONT FACE="Symbol">p</FONT> /4.</P>

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