document.write( "Question 1028882: Solve by completing the square\r
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document.write( "A ball is thrown upward at an initial velocity of 26 ft/sec^2 from a height of 3 ft. Find the approximate length of time(to the nearest tenth of a second) it will take the ball to return to a height of 5 ft. \r
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document.write( "USE THIS FORMULA:\r
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document.write( "h(t)= -16t^2 + vt + h\r
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document.write( "Where V(measured in ft/sec^2) and H(initial height) \n" );
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Algebra.Com's Answer #643962 by mananth(16946)![]() ![]() You can put this solution on YOUR website! \r \n" ); document.write( "\n" ); document.write( "There are 2 times when the ball's height is 5 ft. \n" ); document.write( " On the way up and on the way down. \n" ); document.write( " So if we plug in 5 for h, we should get 2 values for t \n" ); document.write( " 5 = 3 +26t -16t^2 \r \n" ); document.write( "\n" ); document.write( "16t^2-26t+2=0\r \n" ); document.write( "\n" ); document.write( "Find the roots of the equation by quadratic formula \n" ); document.write( " \n" ); document.write( "a= 16 b= -26 c= 2 \n" ); document.write( " \n" ); document.write( "b^2-4ac= 676 - 128 \n" ); document.write( "b^2-4ac= 548 \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "x1=( 26 + 23.41 )/ 32 \n" ); document.write( "x1= 1.54 \n" ); document.write( "x2=( 26 -23.41 ) / 32 \n" ); document.write( "x2= 0.08 \r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "t= 0.08 OR 1.54 \n" ); document.write( " |