document.write( "Question 120158: an object is dropped from an initial height of s feet. the object's height at any time, t in seconds, is given by h=-16t(squared)+s. How long does it take for an object dropped from 300 feet to hit the ground? \n" ); document.write( "
Algebra.Com's Answer #88090 by solver91311(24713)\"\" \"About 
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The initial height, s, is 300 feet, and the final height is 0, because you are concerned with the time it takes to hit the ground.\r
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\n" ); document.write( "\n" ); document.write( "So, your equation becomes:\r
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\n" ); document.write( "\n" ); document.write( "\"-16t%5E2%2B300=0\"\r
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\n" ); document.write( "\n" ); document.write( "Add -300 to both sides and then multiply by -1:\r
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\n" ); document.write( "\n" ); document.write( "\"16t%5E2=300\"\r
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\n" ); document.write( "\n" ); document.write( "Take the square root of both sides:\r
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\n" ); document.write( "\n" ); document.write( "\"4t=sqrt%28300%29\" (you can ignore the negative square root because you are looking for a positive time value)\r
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\n" ); document.write( "\n" ); document.write( "Divide by 4:\r
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\n" ); document.write( "\n" ); document.write( "\"t=sqrt%2875%29\"\r
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\n" ); document.write( "\n" ); document.write( "But \"75=3%2A25\", so \"sqrt%2875%29=5%2Asqrt%283%29\"\r
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\n" ); document.write( "\n" ); document.write( "Hence, \"t=5%2Asqrt%283%29\" seconds.\r
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\n" ); document.write( "\n" ); document.write( "Hope this helps,
\n" ); document.write( "John\r
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