SOLUTION: AN OBJECT is projected vertically upward from the top of a building with an initial velocity of 144 ft./sec. its distance in feet above the ground after t secods is given by the eq

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: AN OBJECT is projected vertically upward from the top of a building with an initial velocity of 144 ft./sec. its distance in feet above the ground after t secods is given by the eq      Log On


   

Question 873162: AN OBJECT is projected vertically upward from the top of a building with an initial velocity of 144 ft./sec. its distance in feet above the ground after t secods is given by the equation s(t)=16t^2+144t+93
solve for t
Answer by mananth(16946) About Me  (Show Source):
You can put this solution on YOUR website!
I think it should be -16x^2
At two times s= 0
at start and when it touches the ground after a time t
equate y=0
and get two values for t