SOLUTION: A BALL THROWN UPWARD WITH AN INITIAL VELOCITY OF 48 FT/SEC FROM A HEIGHT 864 FT. ITS HEIGHT S, IN FEET, AFTER T SECONDS IS GIVEN BY S= -16T^2 +48T+864. AFTER HOW LONG WILL THE BA

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: A BALL THROWN UPWARD WITH AN INITIAL VELOCITY OF 48 FT/SEC FROM A HEIGHT 864 FT. ITS HEIGHT S, IN FEET, AFTER T SECONDS IS GIVEN BY S= -16T^2 +48T+864. AFTER HOW LONG WILL THE BA      Log On


   

Question 974698: A BALL THROWN UPWARD WITH AN INITIAL VELOCITY OF 48 FT/SEC FROM A HEIGHT 864 FT. ITS HEIGHT S, IN FEET, AFTER T SECONDS IS GIVEN BY S= -16T^2 +48T+864. AFTER HOW LONG WILL THE BALL REACH THE GROUND?
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
A BALL THROWN UPWARD WITH AN INITIAL VELOCITY OF 48 FT/SEC FROM A HEIGHT 864 FT. ITS HEIGHT S, IN FEET, AFTER T SECONDS IS GIVEN BY S= -16T^2 +48T+864. AFTER HOW LONG WILL THE BALL REACH THE GROUND?
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S= -16T^2 +48T+864 = 0
S = 0 at impact
Solve for t
Ignore the negative solution.