document.write( "Question 1090948: Please help me solve this problem below.\r
\n" ); document.write( "\n" ); document.write( "An object is launched into the air from a ledge 6 feet off the ground at initial vertical velocity of 96 feet per second. Its height H, in feet, at t second is given by the equation H = 16t^2+96t+16. Find all the times t that the object is at a height of 160 feet off the ground. \n" ); document.write( "

Algebra.Com's Answer #705387 by ikleyn(52781) $\"\"$ $\"About$

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document.write( "    When you consider the problems like this one, the correct form of the equation for the height is \r\n" );
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document.write( "    H(t) = -16t^2 + 96t + 16.\r\n" );
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document.write( "    Notice the sign \"-\" (minus) at the quadratic term.\r\n" );
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document.write( "    Your writing in the post was incorrect.\r\n" );
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\n" ); document.write( "\n" ); document.write( "Second, all you need to do is to solve the equation H(t) = 0, or, which is the same,\r
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\n" ); document.write( "\n" ); document.write( " $\"-16t%5E2+%2B+96t+%2B+16\"$ = 0.\r
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document.write( "For it, divide both sides by 16. You will get\r\n" );
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document.write( "-t^2 + 6t + 1 = 0,    or, which is simpler (and equivalent !)\r\n" );
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document.write( "t^2 -6t -1 = 0.\r\n" );
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document.write( " =  =  = .\r\n" );
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document.write( "Only positive root makes sense.\r\n" );
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document.write( "Answer.  t =  seconds.\r\n" );
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\n" ); document.write( "\n" ); document.write( "Solved.\r
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