document.write( "Question 1152646: An object is launched straight up into the air at an initial velocity of 64 feet per second. It is launched from an initial height of 6 feet off the ground. Find all times t that the object is at a height of 54 feet off the ground.\r
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Algebra.Com's Answer #774690 by ankor@dixie-net.com(22740)![]() ![]() You can put this solution on YOUR website! An object is launched straight up into the air at an initial velocity of 64 feet per second. \n" ); document.write( " It is launched from an initial height of 6 feet off the ground. \n" ); document.write( " Find all times t that the object is at a height of 54 feet off the ground. \n" ); document.write( ": \n" ); document.write( "Using the formula h = -16t^2 + vt + c, where \n" ); document.write( "h = the height above ground at t time \n" ); document.write( "t = time since initially launched \n" ); document.write( "v = initial velocity \n" ); document.write( "c = height above ground at launch \n" ); document.write( ": \n" ); document.write( "h = 54 \n" ); document.write( "-16t^2 + 64t + 6 = 54 \n" ); document.write( "-16t^2 + 64t + 6 - 54 = 0 \n" ); document.write( "-16t^2 + 64t - 48 = 0 \n" ); document.write( "simplify, divide by -16 \n" ); document.write( "t^2 - 4t + 3 = 0 \n" ); document.write( "factors to \n" ); document.write( "(t-3)(t-1) = 0 \n" ); document.write( "two solutions \n" ); document.write( "t = 1 sec at 54 ft on the way up \n" ); document.write( "and \n" ); document.write( "t = 3 sec at 54 ft on the way down \n" ); document.write( " |