document.write( "Question 461487: at 9:45 am Maggie threw a ball upwards while standing on a platform 55 ft off of the ground. The trajectory after t seconds follows the equation: h(t)=-0.6t^2 + 72t + 55
\n" ); document.write( "what will be the maximum height of the ball
\n" ); document.write( "how long will it take the ball to reach its maximum height
\n" ); document.write( "at what time will the ball hit the ground \n" ); document.write( "
Algebra.Com's Answer #316476 by stanbon(75887)\"\" \"About 
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at 9:45 am Maggie threw a ball upwards while standing on a platform 55 ft off of the ground. The trajectory after t seconds follows the equation:
\n" ); document.write( "h(t) = -0.6t^2 + 72t + 55
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\n" ); document.write( "What will be the maximum height of the ball?
\n" ); document.write( "max occurs when t = -b/(2a) = -72/(2*-0.6) = -72/(-1.2) = 6 seconds
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\n" ); document.write( "how long will it take the ball to reach its maximum height? 6 seconds
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\n" ); document.write( "at what time will the ball hit the ground
\n" ); document.write( "Solve -0.6t^2+72t+55 = 0
\n" ); document.write( "Use the quadratic formula:
\n" ); document.write( "t = [-72 +- sqrt(72^2 - 4*-0.6*55)]/(2*-0.6)
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\n" ); document.write( "Positive solution:
\n" ); document.write( "t = 120.76 seconds ~ 2 min
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\n" ); document.write( "the ball will hit the ground at 9:47\r
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