document.write( "Question 1132492: an arrow is shot vertically upward with a speed of 288ft/s, and 3.00 s later another is shot up at a speed of 240 ft/s. will they meet? If so, where? \n" ); document.write( "
Algebra.Com's Answer #749533 by Alan3354(69443)    You can put this solution on YOUR website! arrow is shot vertically upward with a speed of 288ft/s, and 3.00 s later another is shot up at a speed of 240 ft/s. will they meet? If so, where? \n" ); document.write( "=============== \n" ); document.write( "Using 32 ft/sec/sec for gravity: \n" ); document.write( "---- \n" ); document.write( " will they meet? \n" ); document.write( "The 1st arrows height h(t) = -16t^2 + 288t \n" ); document.write( "At t = 3 seconds, h(3) = -16*9 + 288*3 = 720 feet \n" ); document.write( "----- \n" ); document.write( "Its speed = 288 - 3*32 = 192 ft/sec \n" ); document.write( "From that point, its height = -16t^2 + 192t + 720 \n" ); document.write( "======================= \n" ); document.write( "The 2nd arrows height = -16t^2 + 240t \n" ); document.write( "When the 2nd arrow is launched, the 1st arrows height = -16t^2 + 192t + 720 \n" ); document.write( "Find where the 2 heights are equal. \n" ); document.write( "-16t^2 + 192t + 720 = -16t^2 + 240t \n" ); document.write( "48t = 720 \n" ); document.write( "t = 15 seconds --- that's 15 seconds from the 2nd launch. \n" ); document.write( "--------------- \n" ); document.write( "For the 1st arrow: h = -16*18^2 + 288*18 = 0 feet AGL \n" ); document.write( "For the 2nd arrow: h = -16*15^2 + 240*15 = 0 \n" ); document.write( "========== \n" ); document.write( "They meet at impact, 18 seconds after the 1st launch \n" ); document.write( " \n" ); document.write( "  |