document.write( "Question 1087593: When an object is dropped from a height of 20.0 m above the surface of Planet Z, it will fall
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Algebra.Com's Answer #701896 by ikleyn(52781)\"\" \"About 
You can put this solution on YOUR website!
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document.write( "Let  \"a\" be the unknown acceleration under the question, in m/s^2.\r\n" );
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document.write( "Then your equation is\r\n" );
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document.write( "\"%28a%2A2%5E2%29%2F2\" - \"%28a%2A1%5E2%29%2F2\" = 5.   (1)\r\n" );
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document.write( "The term  \"%28a%2A2%5E2%29%2F2\" is the distance that freely falling body passes in two seconds.  (Remember the formula S = \"%28at%5E2%29%2F2\" from Physics ?)\r\n" );
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document.write( "The term  \"%28a%2A1%5E2%29%2F2\" is the distance that freely falling body passes in the first second.\r\n" );
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document.write( "The difference is what you are given.\r\n" );
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document.write( "From the equation (1) you have \r\n" );
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document.write( "4a - a = 2*5,   or\r\n" );
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document.write( "3a = 10,   which gives you  a = \"10%2F3\" = \"3\"\"1%2F3\" m/s^2.\r\n" );
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document.write( "The given value of the height 20 m  IS NOT RELEVANT  to the solution.\r\n" );
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document.write( "We can use it only to check whether our body will be still falling during 2 seconds:\r\n" );
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document.write( "\"%28a%2At%5E2%29%2F2\" = \"%28%28%2810%2F3%29%29%2A2%5E2%29%2F2\" = \"20%2F3\" = 7 m.\r\n" );
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document.write( "Yes, it still be falling.\r\n" );
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document.write( "So, we solved the problem correctly.\r\n" );
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document.write( "Answer.  The acceleration near the surface of the planet is \"10%2F3\" = \"3\"\"1%2F3\" m/s^2.\r\n" );
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