document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=176994>Question 176994</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The path of a falling object is given by the function s=-16t^+v0t+s0 where the v0 represents the initial velocity in ft/sec ans s0represents the initial height. The variable t is time in seconds, and s is the height of the object in feet. if a rock is thrown upward with the initial velocity of 32 feet per secound from the top of a 40 foot building a) write the height equation b)how high is the rock after 5.0 seconds,after how many seconds will the rock reach maximum height and what is the height?  \r<BR>\n" );
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document.write( "I have been stuck on this problem for days I do not know where it is from it is a problem given by our teacher and I have no clue where it is from and that is the truth. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #132038 by </B><A HREF=/tutors/aboutme.mpl?userid=solver91311><B>solver91311(24713)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=solver91311><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=solver91311><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=132038\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=s%28t%29+=+-16t%5E2+%2B+v%5Bo%5Dt+%2B+s%5Bo%5D ALIGN=MIDDLE  ALT=\"s%28t%29+=+-16t%5E2+%2B+v%5Bo%5Dt+%2B+s%5Bo%5D\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "The initial velocity <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=v%5Bo%5D+=+32 ALIGN=MIDDLE  ALT=\"v%5Bo%5D+=+32\"   DISABLED_style= \"max-width:90%; height:auto;\" > is a given (\"...thrown upward with the initial velocity of 32 feet per secound[sic]...\")\r<BR>\n" );
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document.write( "The initial height <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=s%5Bo%5D+=+40 ALIGN=MIDDLE  ALT=\"s%5Bo%5D+=+40\"   DISABLED_style= \"max-width:90%; height:auto;\" > is a given (\"...from the top of a 40 foot building...\")\r<BR>\n" );
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document.write( "So just just substitute:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=s%28t%29+=+-16t%5E2+%2B+32t+%2B+40 ALIGN=MIDDLE  ALT=\"s%28t%29+=+-16t%5E2+%2B+32t+%2B+40\"   DISABLED_style= \"max-width:90%; height:auto;\" > \r<BR>\n" );
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document.write( "Now, the question is what is the height at 5 seconds.  <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=s%28t%29 ALIGN=MIDDLE  ALT=\"s%28t%29\"   DISABLED_style= \"max-width:90%; height:auto;\" > is the height function with respect to time, so just substitute 5 for t and do the arithmetic.\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=s%285%29+=+-16%285%29%5E2+%2B+32%285%29+%2B+40 ALIGN=MIDDLE  ALT=\"s%285%29+=+-16%285%29%5E2+%2B+32%285%29+%2B+40\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=s%285%29+=+-400+%2B+160+%2B+40+=+-200 ALIGN=MIDDLE  ALT=\"s%285%29+=+-400+%2B+160+%2B+40+=+-200\"   DISABLED_style= \"max-width:90%; height:auto;\" > which means that the rock is 200 feet <i>below</i> the ground.  An absurd result to be sure.  Double check the problem you were given because I suspect that the question was to determine the height at 0.5 seconds instead of the 5.0 seconds that you stated.\r<BR>\n" );
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document.write( "The second part of your problem needs you to recognize that the height function is a parabola and that it is concave down because the lead coefficient is negative.  That means the vertex of the parabola is a maximum.\r<BR>\n" );
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document.write( "For any parabola of the form <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=f%28x%29+=+ax%5E2+%2B+bx+%2B+c ALIGN=MIDDLE  ALT=\"f%28x%29+=+ax%5E2+%2B+bx+%2B+c\"   DISABLED_style= \"max-width:90%; height:auto;\" >, the x-coordinate of the vertex is at <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x+=+-b%2F2a ALIGN=MIDDLE  ALT=\"x+=+-b%2F2a\"   DISABLED_style= \"max-width:90%; height:auto;\" >.\r<BR>\n" );
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document.write( "For your problem, <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=a+=+-16 ALIGN=MIDDLE  ALT=\"a+=+-16\"   DISABLED_style= \"max-width:90%; height:auto;\" > and <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=b+=+32 ALIGN=MIDDLE  ALT=\"b+=+32\"   DISABLED_style= \"max-width:90%; height:auto;\" >, so the maximum height is reached at time <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=-32%2F2%28-16%29+=+1 ALIGN=MIDDLE  ALT=\"-32%2F2%28-16%29+=+1\"   DISABLED_style= \"max-width:90%; height:auto;\" > second.\r<BR>\n" );
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document.write( "Calculate <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=s%281%29+=+-16%281%29%5E2+%2B+32%281%29+%2B+40 ALIGN=MIDDLE  ALT=\"s%281%29+=+-16%281%29%5E2+%2B+32%281%29+%2B+40\"   DISABLED_style= \"max-width:90%; height:auto;\" > to get the actual maximum height reached. </SPAN>\n" );
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