document.write( "Question 195469: can you help me solve these problem?
\n" ); document.write( "An arrow is shot into the air with an upward velocity of 48 ft.per second from a hill 32 ft high.The height h of the arrow can be found using the model h=32 +48t -16t^ where t is the time in seconds.How many seconds after release will the arrow be 64 ft. above the ground?I will appreciate very much if you can help me.
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Algebra.Com's Answer #146594 by vleith(2983)\"\" \"About 
You can put this solution on YOUR website!
\"h=32+%2B48t+-16t%5E2\"
\n" ); document.write( "At what time is the height 64?
\n" ); document.write( "\"64+=+32+%2B48t+-16t%5E2\"
\n" ); document.write( "\"0+=+-16t%5E2+%2B+48t+-+32\"
\n" ); document.write( "\"0+=+t%5E2+-3t+%2B+2\"
\n" ); document.write( "\"0+=+%28t-2%29%28t-1%29\"
\n" ); document.write( "t = 1 and t =2\r
\n" ); document.write( "\n" ); document.write( "It is at 64 feet in two places. First at 1 second in, then again at 2 seconds.
\n" ); document.write( "Here is the graph of the arc. Note that the part of the graph to left of time=0, does not apply.
\n" ); document.write( "\"graph%28400%2C400%2C-1%2C4%2C-10%2C90%2C-16x%5E2+%2B+48x+%2B+32%29\"
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