document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=134025>Question 134025</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The formula for the height, h metres, of an object propelled into the air is<BR>\n" );
document.write( "h = -1/2g t squared ( only the t is squared) + vt + h<BR>\n" );
document.write( "where g represents the acceleration due to gravity, which is about 9.8 m/s squared on the Earch. t seconds represents the time, v metres oer secibd reoresents the initial velocity of 34.3 m/s and is launched 2.1 m above the ground. <BR>\n" );
document.write( "a) What is the maximum height, in metres, reached by the projectile?<BR>\n" );
document.write( "b) How many seconds after launch does the projective reach its 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 #98014 by </B><A HREF=/tutors/aboutme.mpl?userid=scott8148><B>scott8148(6628)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=scott8148><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=scott8148><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=98014\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> b) the maximim value is on the axis of symmetry of the parabola<BR>\n" );
document.write( "__ t=-b/2a __ t=-34.3/-9.8 __ t=3.5\r<BR>\n" );
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document.write( "a) h=-4.9t^2+34.3t+2.1 __ substituting __ h=-4.9(3.5)^2+34.3(3.5)+2.1 __ h=62.125 </SPAN>\n" );
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