document.write( "Question 1040257: A body is projected downward at an angle of 30 tp the horizontal with a velocity of 9.8m/s ftom the top of the tower of 29.4m high .how long will it take before striking the ground?
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Algebra.Com's Answer #655049 by ikleyn(52858)\"\" \"About 
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document.write( "The body moves uniformly at the constant speed in horizontal direction and at the constant acceleration \r\n" );
document.write( "g = \"9.8\" \"m%2Fs%5E2\" downward in vertical direction.\r\n" );
document.write( "     (! Do not miss the initial velocity 9.8 \"m%2Fs\" with the gravity acceleration g = \"9.8\" \"m%2Fs%5E2\" !)\r\n" );
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document.write( "To answer the question, it is enough to consider vertical movement of the body.\r\n" );
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document.write( "Its initial vertical velocity is  v = \"sin%2830%5Eo%29%2A9.8\" = \"%281%2F2%29%2A9.8\" = \"4.9\" \"m%2Fs\" directed downward.\r\n" );
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document.write( "Then an equation for the vertical coordinate of the body h(t) is\r\n" );
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document.write( "h(t) = \"-g%2F2%2At%5E2+%2B+v%2At+%2B+29.4\",\r\n" );
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document.write( "where g = \"9.8\" \"m%2Fs%5E2\"  and  v = \"-4.9\" \"m%2Fs\",  or\r\n" );
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document.write( "h(t) = \"-4.9%2At%5E2+-+4.9%2At+%2B+29.4\".\r\n" );
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document.write( "Here h(t) is the vertical distance from the body to the ground level in meters, t is time in seconds.\r\n" );
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document.write( "To find the time moment when the body strikes the ground, you need to solve a quadratic equation \r\n" );
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document.write( "h(t) = 0,  or  \"-4.9%2At%5E2+-+4.9%2At+%2B+29.4\" = \"0\".\r\n" );
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document.write( "It is the same as \r\n" );
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document.write( "\"4.9%2A%28t%5E2+%2B+t+-+6%29\" = \"0\". \r\n" );
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document.write( "The last equation is equivalent to \r\n" );
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document.write( "\"t%5E2+%2B+t+-+6\" = \"0\".\r\n" );
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document.write( "Factor it. It is equivalent to\r\n" );
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document.write( "(t-3)*(t+2) = 0\r\n" );
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document.write( "and has the roots t = 3 and t = -2.\r\n" );
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document.write( "Only positive root is acceptable.\r\n" );
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document.write( "So, the answer is: t = 3.  The body hits the ground in 3 seconds.\r\n" );
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