document.write( "Question 214728: An arrow is shot vertically upward. Three seconds later, it is at a height of 35m.
\n" ); document.write( "What was the initial speed of the arrow?
\n" ); document.write( "How long is the arrow in flight from launch to returning to the initial height? (Assume the arrow falls vertically downward.)
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Algebra.Com's Answer #162192 by ankor@dixie-net.com(22740)\"\" \"About 
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An arrow is shot vertically upward. Three seconds later, it is at a height of 35m.
\n" ); document.write( "What was the initial speed of the arrow?
\n" ); document.write( ":
\n" ); document.write( "Using the equation -4.9t^2 + vt + c = H
\n" ); document.write( "Where
\n" ); document.write( "h = height after t seconds (35 meters)
\n" ); document.write( "t = time in seconds (3)
\n" ); document.write( "-4.9t^2 = the downward pull of gravity
\n" ); document.write( "v = upward velocity
\n" ); document.write( "c = initial height (0 in this problem)
\n" ); document.write( ":
\n" ); document.write( "-4.9(3)^2 + 3v = 35
\n" ); document.write( "-4.9(9) + 3v = 35
\n" ); document.write( "-44.1 + 3v = 35
\n" ); document.write( "3v = 35 + 44.1
\n" ); document.write( "3v = 75.1
\n" ); document.write( "v = \"75.1%2F3\"
\n" ); document.write( "v = 26.37 m/sec initial velocity
\n" ); document.write( ":
\n" ); document.write( "How long is the arrow in flight from launch to returning to the initial height? (Assume the arrow falls vertically downward.)
\n" ); document.write( "-4.9t^2 + 26.37t = 0
\n" ); document.write( "Factor out -4.9t
\n" ); document.write( "-4.9t(t - 5.38) = 0
\n" ); document.write( "Two solutions
\n" ); document.write( "t = 0 sec; arrow on it's way
\n" ); document.write( "t = 5.38 sec when it returns to earth (time of flight)
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