Question 23272: SUPPOSE YOU THROW A BASEBALL STRAIGHT UP AT A VELOCITY OF 64 FEET PER SECOND. A FUNCTION
CAN BE CREATED BY EXPRESSING DISTANCE ABOVE GROUND, S, AS A FUNCTION OF TIME, T. THIS FUNCTION IS: S=-16T^2+^V0^T+^S0
16 REPRESENTS 1/2G, THE GRAVITATIONAL PULL DUE TO GRAVITY (MEASURED IN FEET PER SECOND^2).
^V0 IS THE INITIAL VELOCITY (HOW HARD DO YOU THROW THE OBJECT, MEASURED IN FEET PER SECOND).
^S0 IS THE INITIAL DISTANCE ABOVE GROUND (IN FEET). IF YOU ARE STANDING ON THE GROUND, THEN ^S0=0.
WHAT IS THE FUNCTION THAT DESCIBES THIS PROBLEM?
THE BALL WILL BE HOW HIGH ABOVE GROUND AFTER 1 SECOND?
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! s(t) = -16t^2+64t+0
"t" is number of seconds after you have thrown the ball.
So, let t=1 and solve for s(1)
s(1) = -16+64= 48 ft above the ground.
Cheers,
Stan H.
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