document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1203081>Question 1203081</A>: <I><SPAN STYLE=\"word-break: break-all;\"> If a ball is thrown upward from the top of a building 30m high with an initial velocity of 10m/s, then the height (h) above ground t seconds later will be <BR>\n" );
document.write( "h = 30 + 10t - 5t^2\r<BR>\n" );
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document.write( "During what time interval will the ball be at least 15m above the ground? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #838339 by </B><A HREF=/tutors/aboutme.mpl?userid=Theo><B>Theo(13342)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Theo><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=Theo><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=838339\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> set the equation equal to 15 to get:<BR>\n" );
document.write( "30 + 10t - 5t^2 = 15<BR>\n" );
document.write( "subtract 15 from both sides of the equation to get:<BR>\n" );
document.write( "15 + 10t - 5t^2 = 0<BR>\n" );
document.write( "rearrange the equation in descending order of degree to get:<BR>\n" );
document.write( "-5t^2 + 10t + 15 = 0<BR>\n" );
document.write( "multiply both sides of the equation by -1 to get:<BR>\n" );
document.write( "5t^2 - 10t - 15 = 0<BR>\n" );
document.write( "divide both sides of the equation by 5 to get:<BR>\n" );
document.write( "t^2 - 2t - 3 = 0<BR>\n" );
document.write( "factor the equation to get:<BR>\n" );
document.write( "(t-3) * (t+1) = 0<BR>\n" );
document.write( "solve for t to get:<BR>\n" );
document.write( "t = 3 or t = -1<BR>\n" );
document.write( "t has to be positive so t = 3.<BR>\n" );
document.write( "when t = 3, -5t^2 + 10t + 15 becomes -45 + 30 + 15 whih becomes 0, as it should.<BR>\n" );
document.write( "go back to the original equqtion of h = 30 + 10t - 5t^2 and replace t with 3 to get:<BR>\n" );
document.write( "h = 30 + 30 - 45 = 15.<BR>\n" );
document.write( "that's the height when t = 3.<BR>\n" );
document.write( "that's your solution.<BR>\n" );
document.write( "here's a graph of the result.\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" src = \"http://theo.x10hosting.com/2023/071901.jpg\">\r<BR>\n" );
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document.write( "in the graph, y takes the place of h and x takes the place of t.\r<BR>\n" );
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document.write( "the value of h will be at least 15 feet above the ground from t = 0 to t = 3.<BR>\n" );
document.write( "in the graph, that's from x = 0 to x = 3.\r<BR>\n" );
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document.write( "while the graph does show negaive values of t, that is not part of the domain since t has to be greater than or equal to 0 (time is not negative).\r<BR>\n" );
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document.write( "to see that clearer, just draw a horizontal line at y = 15 on the graph.<BR>\n" );
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