document.write( "Question 35777: the height h, of a ball above ground t seconds after it is thrown vertically upwards can be approximated by the formula h=vt-5t squared, wher v is the intial speed with which the ball is released in meters per second.
\n" ); document.write( "at what two times will a ball be 105m above ground if it is thrown vertically upwards with an initial speed of 50m/s \n" ); document.write( "

Algebra.Com's Answer #21891 by Earlsdon(6294) $\"\"$ $\"About$

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$\"h+=+vt-5t%5E2\"$ \r
\n" ); document.write( "\n" ); document.write( "Substitute v = 50 and h = 105 and solve for t.
\n" ); document.write( " $\"105+=+50t+-+5t%5E2\"$ Subsubtract 105 from both sides and simplify.
\n" ); document.write( " $\"5t%5E2+-+50t+%2B+105+=+0\"$ Solve this quadratic equation by factoring.
\n" ); document.write( " $\"5%28t%5E2+-+10t+%2B+21%29+=+0\"$ Apply the zero products principle.
\n" ); document.write( " $\"t%5E2+-+10t+%2B+21+=+0\"$ Factor.
\n" ); document.write( " $\"%28t+-+3%29%28t+-+7%29+=+0\"$ Apply the zero products principle again.
\n" ); document.write( " $\"t+-+3+=+0\"$ and $\"t+-+7+=+0\"$
\n" ); document.write( "If $\"t+-+3+=+0\"$ then $\"t+=+3\"$
\n" ); document.write( "If $\"t+-+7+=+0\"$ then $\"t+=+7\"$ \r
\n" ); document.write( "\n" ); document.write( "The two times are:
\n" ); document.write( "t = 3 seconds.
\n" ); document.write( "t = 7 seconds.\r
\n" ); document.write( "\n" ); document.write( "You check these solutions by substituting into the original equation $\"h+=+50t-5t%5E2\"$ and solving for h. \n" ); document.write( "

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