SOLUTION: a ball is thrown vertically upward with an initial speed of 48 ft/s. its height, in feet, after t seconds is given by h=48t-16t^2. how high does the ball go?

Question 860825: a ball is thrown vertically upward with an initial speed of 48 ft/s. its height, in feet, after t seconds is given by h=48t-16t^2. how high does the ball go?
Answer by ewatrrr(24785) (Show Source): You can put this solution on YOUR website!

 

Hi,

h = -16t^2 + 48t

h = -16(t-3/2)^2 + 16(9/4)

h = -16(t-3/2)^2 + 36

 36ft is how high the ball goes (reaches that height after 1.5 sec) 

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