document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=64597>Question 64597</A>: <I><SPAN STYLE=\"word-break: break-all;\"> I really, really need help on this one..... The (-32x^2 over 310^2), in the equation below, is a fraction......\r<BR>\n" );
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document.write( "A projectile is fired from a cliff 200 feet above the water at an incline of 45 degrees to the horizontal, with a muzzle velocity of 310 feet per second. The height h of the projectile above the water is given by:  <BR>\n" );
document.write( "h(x)= (-32x^2 over 310^2)+x+200 , where x is the horizontal distance of the projectile from the base of the cliff. Find the maximun height of the projectile. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #46478 by </B><A HREF=/tutors/aboutme.mpl?userid=venugopalramana><B>venugopalramana(3286)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=venugopalramana><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.geocities.com/venugopalramana/FREE_MATHS_ASSISTANCE.html valign=middle><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=46478\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> I really, really need help on this one..... The (-32x^2 over 310^2), in the equation below, is a fraction......<BR>\n" );
document.write( "A projectile is fired from a cliff 200 feet above the water at an incline of 45 degrees to the horizontal, with a muzzle velocity of 310 feet per second. The height h of the projectile above the water is given by:<BR>\n" );
document.write( "h(x)= (-32x^2 over 310^2)+x+200 , where x is the horizontal distance of the projectile from the base of the cliff. Find the maximun height of the projectile.<BR>\n" );
document.write( "H=(-32X^2/310^2)+X+200<BR>\n" );
document.write( "HORIZONTAL COMPONENT OF VELOCITY=310COS(45)=310/SQRT(2)=310SQRT(2)/2=155SQRT(2)=219.2 FT/SEC<BR>\n" );
document.write( "VERTICAL COMPONENT OF VELOCITY =310SIN(45)=155SQRT(2)=219.2 FT/SEC<BR>\n" );
document.write( "THE PROJECTILE WILL GO UP TILL ITS VELOCITY=0 AND THEN FALLS DOWN.<BR>\n" );
document.write( "FORMULA IS V=U+AT...<BR>\n" );
document.write( "WHERE U = INITIAL VELOCITY IN THE VERTICAL DIRECTION=155SQRT(2) FT/SEC<BR>\n" );
document.write( "V=FINAL VELOCITY =0 FT/SEC<BR>\n" );
document.write( "A=ACCELERATION DUE TO GRAVITY = -32 FT/SEC^2 <BR>\n" );
document.write( "T= TIME OF TRAVEL IN SEC.<BR>\n" );
document.write( "0=155SQRT(2)-32T<BR>\n" );
document.write( "T=155SQRT(2)/32=6.85 SEC<BR>\n" );
document.write( "HORIZINTAL TRAVEL IN THIS TIME OF 6.85 SEC = 155SQRT(2)*6.85=1501.5<BR>\n" );
document.write( "H=-(32*1501.5^2/310^2)+1501.5+200=950.5 FEET.ABOVE THE WATER LEVEL OR 750.5 FEET ABOVE THE CLIFF LEVEL.<BR>\n" );
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document.write( "PLEASE NOTE THAT THERE ARE SEVERAL WAYS TO DO THIS PROBLEM,USING FORMULAE,CALCULUS ETC.IF I KNOW YOUR COURSE OF STUDY AND BACK GROUND ,I CAN GIVE THE METHOD BEST SUITED TO YOU. </SPAN>\n" );
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