document.write( "Question 950715: If an object is projected vertically upward from ground level with an initial velocity of 280 ft/sec, then its distance s above the ground after t seconds is given by
\n" ); document.write( "s = −14t2 + 280t.
\n" ); document.write( " For what values of t will the object be more than 1050 feet above the ground?\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "______ < t < ______ \n" ); document.write( "

Algebra.Com's Answer #580566 by josmiceli(19441) $\"\"$ $\"About$

You can put this solution on YOUR website!
$\"+s+%3E+1050+\"$
\n" ); document.write( " $\"+s+=+-14t%5E2+%2B+280t+\"$
\n" ); document.write( "----------------------
\n" ); document.write( "Set $\"+s+=+1050+\"$ ft
\n" ); document.write( " $\"+1050+=+-14t%5E2+%2B+280t+\"$
\n" ); document.write( " $\"+-14t%5E2+%2B+280t+-+1050+=+0+\"$
\n" ); document.write( " $\"+-7t%5E2+%2B+140t+-+525+=+0+\"$
\n" ); document.write( " $\"+-t%5E2+%2B+20t+-+75+=+0+\"$
\n" ); document.write( "I notice that $\"+75+=+15%2A5+\"$ , so I
\n" ); document.write( "can say:
\n" ); document.write( " $\"+%28+-t+%2B+15+%29%2A%28+t+-+5+%29+=+0+\"$
\n" ); document.write( " $\"+-t+%2B+15+=+0+\"$
\n" ); document.write( " $\"+t+=+15+\"$
\n" ); document.write( "and
\n" ); document.write( " $\"+t+-+5+=+0+\"$
\n" ); document.write( " $\"+t+=+5+\"$
\n" ); document.write( "-------------
\n" ); document.write( " $\"+s+%3E+1050+\"$ ft for
\n" ); document.write( " $\"+t+%3E+5+\"$ sec
\n" ); document.write( " $\"+t+%3C+15+\"$ sec
\n" ); document.write( "---------------
\n" ); document.write( " $\"+5+%3C+t+%3C+15+\"$ sec
\n" ); document.write( "------------------
\n" ); document.write( "Here's the plot:
\n" ); document.write( " $\"+graph%28+400%2C+400%2C+-5%2C+25%2C+-300%2C+1500%2C+-14x%5E2+%2B+280x%2C+1050+%29+\"$ \r
\n" ); document.write( "\n" ); document.write( " \n" ); document.write( "

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