document.write( "Question 23272: SUPPOSE YOU THROW A BASEBALL STRAIGHT UP AT A VELOCITY OF 64 FEET PER SECOND. A FUNCTION
\n" ); document.write( "CAN BE CREATED BY EXPRESSING DISTANCE ABOVE GROUND, S, AS A FUNCTION OF TIME, T. THIS FUNCTION IS: S=-16T^2+^V0^T+^S0\r
\n" ); document.write( "\n" ); document.write( "16 REPRESENTS 1/2G, THE GRAVITATIONAL PULL DUE TO GRAVITY (MEASURED IN FEET PER SECOND^2).
\n" ); document.write( "^V0 IS THE INITIAL VELOCITY (HOW HARD DO YOU THROW THE OBJECT, MEASURED IN FEET PER SECOND).
\n" ); document.write( "^S0 IS THE INITIAL DISTANCE ABOVE GROUND (IN FEET). IF YOU ARE STANDING ON THE GROUND, THEN ^S0=0.
\n" ); document.write( "WHAT IS THE FUNCTION THAT DESCIBES THIS PROBLEM?
\n" ); document.write( "THE BALL WILL BE HOW HIGH ABOVE GROUND AFTER 1 SECOND?
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Algebra.Com's Answer #12021 by stanbon(75887)\"\" \"About 
You can put this solution on YOUR website!
s(t) = -16t^2+64t+0
\n" ); document.write( "\"t\" is number of seconds after you have thrown the ball.
\n" ); document.write( "So, let t=1 and solve for s(1)
\n" ); document.write( "s(1) = -16+64= 48 ft above the ground.\r
\n" ); document.write( "\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H.
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