document.write( "Question 121755: The equation h=-16t^2 + 112t gives the height of an arrow, shot upward from the ground with an initial velocity of 112ft/s, where t is the time after the arrow leaves the ground . Find the time it takes for the arrow to reach a height of 180 ft?\r
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\n" ); document.write( "\n" ); document.write( "Not sure if I'm doing the problem right but just double checking.
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Algebra.Com's Answer #89400 by jim_thompson5910(35256)\"\" \"About 
You can put this solution on YOUR website!
\"h+=+-16t%5E2+%2B+112t\" Start with the given equation\r
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\n" ); document.write( "\n" ); document.write( "\"180+=+-16t%5E2+%2B+112t\" Plug in \"h=180\"\r
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\n" ); document.write( "\n" ); document.write( "\"16t%5E2-112t%2B180=0\" Get everything to the left side\r
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\n" ); document.write( "\n" ); document.write( "Let's use the quadratic formula to solve for t:\r
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\n" ); document.write( "\n" ); document.write( "Starting with the general quadratic\r
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\n" ); document.write( "\n" ); document.write( "\"at%5E2%2Bbt%2Bc=0\"\r
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\n" ); document.write( "\n" ); document.write( "the general solution using the quadratic equation is:\r
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\n" ); document.write( "\n" ); document.write( "\"t+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29\"\r
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\n" ); document.write( "\n" ); document.write( "So lets solve \"16%2At%5E2-112%2At%2B180=0\" ( notice \"a=16\", \"b=-112\", and \"c=180\")\r
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\n" ); document.write( "\n" ); document.write( "\"t+=+%28--112+%2B-+sqrt%28+%28-112%29%5E2-4%2A16%2A180+%29%29%2F%282%2A16%29\" Plug in a=16, b=-112, and c=180\r
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\n" ); document.write( "\n" ); document.write( "\"t+=+%28112+%2B-+sqrt%28+%28-112%29%5E2-4%2A16%2A180+%29%29%2F%282%2A16%29\" Negate -112 to get 112\r
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\n" ); document.write( "\n" ); document.write( "\"t+=+%28112+%2B-+sqrt%28+12544-4%2A16%2A180+%29%29%2F%282%2A16%29\" Square -112 to get 12544 (note: remember when you square -112, you must square the negative as well. This is because \"%28-112%29%5E2=-112%2A-112=12544\".)\r
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\n" ); document.write( "\n" ); document.write( "\"t+=+%28112+%2B-+sqrt%28+12544%2B-11520+%29%29%2F%282%2A16%29\" Multiply \"-4%2A180%2A16\" to get \"-11520\"\r
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\n" ); document.write( "\n" ); document.write( "\"t+=+%28112+%2B-+sqrt%28+1024+%29%29%2F%282%2A16%29\" Combine like terms in the radicand (everything under the square root)\r
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\n" ); document.write( "\n" ); document.write( "\"t+=+%28112+%2B-+32%29%2F%282%2A16%29\" Simplify the square root (note: If you need help with simplifying the square root, check out this solver)\r
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\n" ); document.write( "\n" ); document.write( "\"t+=+%28112+%2B-+32%29%2F32\" Multiply 2 and 16 to get 32\r
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\n" ); document.write( "\n" ); document.write( "So now the expression breaks down into two parts\r
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\n" ); document.write( "\n" ); document.write( "\"t+=+%28112+%2B+32%29%2F32\" or \"t+=+%28112+-+32%29%2F32\"\r
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\n" ); document.write( "\n" ); document.write( "Lets look at the first part:\r
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\n" ); document.write( "\n" ); document.write( "\"x=%28112+%2B+32%29%2F32\"\r
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\n" ); document.write( "\n" ); document.write( "\"t=144%2F32\" Add the terms in the numerator\r
\n" ); document.write( "\n" ); document.write( "\"t=9%2F2\" Divide\r
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\n" ); document.write( "\n" ); document.write( "So one answer is\r
\n" ); document.write( "\n" ); document.write( "\"t=9%2F2\"\r
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\n" ); document.write( "\n" ); document.write( "Now lets look at the second part:\r
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\n" ); document.write( "\n" ); document.write( "\"x=%28112+-+32%29%2F32\"\r
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\n" ); document.write( "\n" ); document.write( "\"t=80%2F32\" Subtract the terms in the numerator\r
\n" ); document.write( "\n" ); document.write( "\"t=5%2F2\" Divide\r
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\n" ); document.write( "\n" ); document.write( "So another answer is\r
\n" ); document.write( "\n" ); document.write( "\"t=5%2F2\"\r
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\n" ); document.write( "\n" ); document.write( "So our solutions are:\r
\n" ); document.write( "\n" ); document.write( "\"t=9%2F2\" or \"t=5%2F2\" (which are \"t=4.5\" or \"t=2.5\" respectively in decimal form)\r
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\n" ); document.write( "\n" ); document.write( "So it takes 2.5 seconds for the arrow to climb to 180 ft and then 4.5 seconds for the arrow to fall back to 180 ft
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