SOLUTION: An object is thrown upward with a velocity of 48.0 m/s. When will it be 25.0 m above its initial position? Note that g = 9.8 m/s2. Round to three decimal places. Give answers in

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equation Customizable Word Problems -> SOLUTION: An object is thrown upward with a velocity of 48.0 m/s. When will it be 25.0 m above its initial position? Note that g = 9.8 m/s2. Round to three decimal places. Give answers in      Log On


   

Question 1060618: An object is thrown upward with a velocity of 48.0 m/s. When will it be 25.0 m above its initial position?
Note that g = 9.8 m/s2. Round to three decimal places. Give answers in increasing order.
Answer by Boreal(15235) About Me  (Show Source):
You can put this solution on YOUR website!
height=-9.8t^2+48t
25=-9.8t^2+48t
9.8t^2-48t+25=0, moving and checking signs.
t=(1/19.6)(48+/-sqrt(2304-980); sqrt (1324)=36.387
t=(11.613/19.6)=0.593 sec
t=84.387/19.6=4.305 sec
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