document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1178109>Question 1178109</A>: <I><SPAN STYLE=\"word-break: break-all;\"> An object moving vertically is at the given heights at the specified times. Find the position equation s =1/2at^2 + v0t + s0 for the object.<BR>\n" );
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document.write( "t = 1 second, s = 156 feet\r<BR>\n" );
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document.write( "t = 2 seconds, s = 108 feet\r<BR>\n" );
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document.write( "t = 3 seconds, s = 28 feet </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #807275 by </B><A HREF=/tutors/aboutme.mpl?userid=Boreal><B>Boreal(15235)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Boreal><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=Boreal><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=807275\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> s=-16t^2+vt+s0<BR>\n" );
document.write( "156=-16+v+s0; v+s0=172<BR>\n" );
document.write( "108=-64+2v+s0; 2v+s0=172; therefore v=0 and s0=172<BR>\n" );
document.write( "28=-144+3v+s0; 3v+s0=172<BR>\n" );
document.write( "The equation is s(g)=-(1/2)at^2+172<BR>\n" );
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document.write( "It is dropped from a height of 172 feet.<BR>\n" );
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