document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=177100>Question 177100</A>This question is from textbook <A HREF=http://www.amazon.com/exec/obidos/ASIN/0131953648/algebrahomeworkh>Blitzer College Algebra</A><BR>\n" );
document.write( ": <I><SPAN STYLE=\"word-break: break-all;\"> If a rock is thrown upward with an initial velocity of 32 feet per second from the top of a 40-story building,write the height equation using this information.<BR>\n" );
document.write( "How high is the rock after 0.5 seconds? How many seconds will the rock reach maximum height? What is the maximum height? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #724900 by </B><A HREF=/tutors/aboutme.mpl?userid=amalm06><B>amalm06(224)</B></A><img src=\"/images/stars/stars-09.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=amalm06><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=amalm06><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=724900\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> The problem can be solved using momentum and energy considerations. The net force acting on the rock causes the velocity of the rock to change. This is given by the following impulse-momentum equation:\r<BR>\n" );
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document.write( "F&#916;t=m(vf-v0)\r<BR>\n" );
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document.write( "Since gravity is the only force that acts on the rock:\r<BR>\n" );
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document.write( "mg&#916;t=m(vf-v0)<BR>\n" );
document.write( " g&#916;t=vf-v0, where g=-32 ft/s^2.\r<BR>\n" );
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document.write( "Then (-32)(&#916;t)=vf-v0\r<BR>\n" );
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document.write( "At maximum height, the velocity of the rock is zero. If you imagine the trajectory of the rock to follow a symmetrical parabolic path, then the slope of the horizontal tangent is zero at maximum height. But this slope is just the first derivative of the position, which is the velocity. \r<BR>\n" );
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document.write( "Then (-32)((&#916;t)=0-32 = -32\r<BR>\n" );
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document.write( "Therefore, t=1. The rock reaches maximum height after 1 second (Answer).\r<BR>\n" );
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document.write( "At maximum height, the rock has only potential energy. Since we are neglecting air resistance, friction does no work on the rock:\r<BR>\n" );
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document.write( "(1/2)(m)(V^2)i=mg(h2-40)<BR>\n" );
document.write( "0.5(v^2)i=g(h2-40)<BR>\n" );
document.write( "(0.5)(32)=(32)(h2-40)<BR>\n" );
document.write( "16=h2-40<BR>\n" );
document.write( "h2=56 ft. The maximum height of the rock is 56 ft (Answer)\r<BR>\n" );
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document.write( "To find the height of the rock after 0.5 seconds, use the following formula:\r<BR>\n" );
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document.write( "&#916;y=Vit + 0.5a(t^2)\r<BR>\n" );
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document.write( "So that y2-40=32t+(0.5)(-32)(t^2)\r<BR>\n" );
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document.write( "        y2=32t-16(t^2)+40\r<BR>\n" );
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document.write( "        y2=-16(t^2)+32t+40\r<BR>\n" );
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document.write( "For t=0.5 seconds, y2=-4+16+40=52 ft (Answer)\r<BR>\n" );
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document.write( " </SPAN>\n" );
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