document.write( "Question 1152706: An astronaut on the moon throws a baseball upward. The astronaut is 6β ft, 6 in.β tall, and the initial velocity of the ball is 30 ft per sec. The height s of the ball in feet is given by the equation s equals -2.7t^2 + 30t + 6.5β, where t is the number of seconds after the ball was thrown. Complete parts a and b.
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document.write( "a. After how many seconds is the ball 12 ft above theβ moon's surface? \n" );
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Algebra.Com's Answer #774786 by ikleyn(52781)![]() ![]() You can put this solution on YOUR website! . \n" ); document.write( " \r\n" ); document.write( "\r\n" ); document.write( "Solve the quadratic equation\r\n" ); document.write( "\r\n" ); document.write( "\r\n" ); document.write( " -2.7t^2 + 30t + 6.5 = 12\r\n" ); document.write( "\r\n" ); document.write( "\r\n" ); document.write( "using the quadratic formula.\r\n" ); document.write( "\r\n" ); document.write( "\r\n" ); document.write( "The two solutions will give you two time moments.\r\n" ); document.write( "\r \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |