document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=49217>Question 49217</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A circus clown at the top of a 60-foot platform is playing catch with another clown on the ground.  The clown on the platform drops a ball at the same time as the one on the ground tosses a ball upward at 80 ft/sec.  For what length of time is the distance between the balls less than or equal to 10 feet?  Hint:  the initial velocity of a ball that is dropped is 0 ft/sec.  The formula for the height S in feet above the earth at a time t seconds for an object projected into the air with an initial velocity of v ft/sec from an initial height of s0 is:<BR>\n" );
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document.write( "S = -16t2 + vt + s0                                                                        </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #32748 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=32748\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> A<BR>\n" );
document.write( "circus clown at the top of a 60-foot platform is<BR>\n" );
document.write( "playing catch with another clown on the ground.<BR>\n" );
document.write( "The clown on the platform drops a ball at the same<BR>\n" );
document.write( "time as the one on the ground tosses a ball upward at<BR>\n" );
document.write( "80 ft/sec.<BR>\n" );
document.write( "For what length of time is the distance between the<BR>\n" );
document.write( "balls less than or equal to 10 feet? Hint: the initial<BR>\n" );
document.write( "velocity of a ball that is dropped is 0 ft/sec. The<BR>\n" );
document.write( "formula for the height S in feet above the earth at a<BR>\n" );
document.write( "time t seconds for an object projected into the air<BR>\n" );
document.write( "with an initial velocity of v ft/sec from an initial<BR>\n" );
document.write( "height of s0 is:<BR>\n" );
document.write( "S = -16t2 + v0t + s0<BR>\n" );
document.write( "1 solutions<BR>\n" );
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document.write( "Answer 17287 by venugopalramana(1619) on 2006-03-18<BR>\n" );
document.write( "11:02:08 (Show Source):<BR>\n" );
document.write( "BASIS..START TIME ..0 SEC.......GROUND LEVEL =0 FT.ALL<BR>\n" );
document.write( "HEIGHTS MEASURED FROM GROUND LEVEL.<BR>\n" );
document.write( "WE HAVE 2 BODIES B1 COMING DOWN FROM TOP TO<BR>\n" );
document.write( "GROUND...SAY.... U TO P.<BR>\n" );
document.write( "B2 GOING UP FROM GROUND UPWARDS......SAY....P TO<BR>\n" );
document.write( "U....(OFCOURSE NOT CLASHING)<BR>\n" );
document.write( "ATTACHMENT:-<BR>\n" );
document.write( "LET B1 COME DOWN FROM U TO R AND B2 GO UP FROM P TO Q<BR>\n" );
document.write( "WHEN THEY ARE AT THE 10 FEET SEPERATION ASKED FOR.THIS<BR>\n" );
document.write( "IS THE ATTACHMENT LET US SAY.LET THIS HAPPEN IN T1<BR>\n" );
document.write( "SECS.FROM START.WE HAVE<BR>\n" );
document.write( "PU=60 FEET...LET PQ=X FT.....QR=10 FT.....HENCE<BR>\n" );
document.write( "PR=X+10 FEET.<BR>\n" );
document.write( "DISTANCE TRAVELLED IN T SECS.FROM START,MEASURED FROM<BR>\n" );
document.write( "GROUND LEVEL IS GIVEN BY<BR>\n" );
document.write( "S=-16T^2+(V0)T+(S0).....WHERE V0 IS INITIAL VELOCITY<BR>\n" );
document.write( "AND S0 IS START DISTANCE FROM GROUND.<BR>\n" );
document.write( "SO FOR B1 COMING DOWN WE HAVE<BR>\n" );
document.write( "X+10=-16T1^2+0*T1+60=-16T1^2+60...............................I<BR>\n" );
document.write( "FOR B2 GOING UP ,WE HAVE...<BR>\n" );
document.write( "X=-16T1^2+80T1+0=-16T1^2+80T1..................................II<BR>\n" );
document.write( "SUBSTITUTING FOR X FROM EQN.II IN EQN.I,WE GET<BR>\n" );
document.write( "-16T1^2+80T1+10=-16T1^2+60<BR>\n" );
document.write( "80T1=50<BR>\n" );
document.write( "T1=50/80=5/8 SECS.<BR>\n" );
document.write( "SO THE ATTACHMENT TAKES PLACE T 5/8 SEC. FROM START.<BR>\n" );
document.write( "DETACHMENT:-<BR>\n" );
document.write( "NOW B1 CONTINUES TO COME DOWN AS B2 GOES UP,CROSSING<BR>\n" );
document.write( "EACH OTHER,WHEN THE DISTANE BETWEEN THEM COMES DOWN<BR>\n" );
document.write( "FROM 10 FT. TO ZERO.AS THEY FURTHER CONTINUE THEIR<BR>\n" );
document.write( "TRAVEL ,THEIR DISTANCE OF SEPERATION INCREASES FROM<BR>\n" );
document.write( "ZERO TO 10 FT.LET US CALL THIS DETACHMENT.LET THIS<BR>\n" );
document.write( "HAPPEN AFTER T2 SECS.FROM START WHEN B1 REACHES SAY S<BR>\n" );
document.write( "AND B2 REACHES T.WE HAVE..<BR>\n" );
document.write( "LET PT=Y AND HENCE PS=Y-10...SO FOR THIS PART<BR>\n" );
document.write( "FOR B1 COMING DOWN WE HAVE..<BR>\n" );
document.write( "Y-10=-16T2^2+0*T2+60=-16T2^2+60...........................III<BR>\n" );
document.write( "FOR B2 GOING UP,WE HAVE..<BR>\n" );
document.write( "Y=-16T2^2+80T2+0=-16T2^2+80T2...........................IV<BR>\n" );
document.write( "SUBSTITUTING FOR Y FROM EQN.III IN EQN.IV..WE GET<BR>\n" );
document.write( "-16T2^2+80T2-10=-16T2^2+60<BR>\n" );
document.write( "80T2=70<BR>\n" );
document.write( "T2=7/8 SECS.<BR>\n" );
document.write( "SO THE DETACHMENT TAKES PLACE AT 7/8 SECS.FROM START.<BR>\n" );
document.write( "SO FOR 7/8-5/8=2/8=1/4 SECS.THE 2 BALLS ARE AT A<BR>\n" );
document.write( "DISTANCE OF 10 FEET OR LESS. </SPAN>\n" );
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