document.write( "Question 892229: A frog jumps straight up off the ground with an initial vertical velocity of 2 feet per second. After how many seconds does the frog land on the ground? Use the vertical motion model. h=-16t^2 + vt + s \n" ); document.write( "

Algebra.Com's Answer #540413 by KMST(5328) $\"\"$ $\"About$

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$\"h=-16t%5E2+%2B+vt+%2B+s\"$ with
\n" ); document.write( " $\"t\"$ =time in seconds,
\n" ); document.write( " $\"h\"$ = height in feet above the ground,
\n" ); document.write( " $\"s\"$ = initial height in feet above the ground (at time zero), and
\n" ); document.write( " $\"v\"$ = initial velocityin feet per second,\r
\n" ); document.write( "\n" ); document.write( "In this case $\"v=2\"$ and $\"s=0\"$ , so
\n" ); document.write( " $\"h=-16t%5E2+%2B+2t+%2B+0\"$ or simply $\"h=-16t%5E2+%2B+2t\"$
\n" ); document.write( "Factoring, we get
\n" ); document.write( " $\"h=%28-16t+%2B+2%29%2At\"$
\n" ); document.write( "When the frog is on the ground, $\"h=0\"$ , and
\n" ); document.write( " $\"0=%28-16t+%2B+2%29%2At\"$ has 2 solutions: $\"system%28t=0%2C%22or%22%2C-16t%2B2=0%29\"$
\n" ); document.write( " $\"t=0\"$ is the time when the frog pushed off the gound.
\n" ); document.write( "The solution to $\"-16t%2B2=0\"$ gives us the time the frog is on the ground again.
\n" ); document.write( " $\"-16t%2B2=0\"$ --> $\"2=16t\"$ --> $\"2%2F16=t\"$ --> $\"highlight%28t=1%2F8%29\"$ or $\"highlight%28t=0.125%29\"$
\n" ); document.write( "The frog lands after $\"1%2F8\"$ of a second (or 0.125 seconds). \n" ); document.write( "

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