SOLUTION: A diver jumps from a 48 ft cliff into the ocean below with the initial velocity of 8 ft per sec.. The height of the diver from the ocean after t sec is given by h = -16t^2 + 8t + 4
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Question 284250: A diver jumps from a 48 ft cliff into the ocean below with the initial velocity of 8 ft per sec.. The height of the diver from the ocean after t sec is given by h = -16t^2 + 8t + 48. When does the diver hit the water?
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
He hits the water when h = 0
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h = -16t^2 + 8t + 48
-16t^2 + 8t + 48 = 0
-2t^2 + t + 3 = 0
(-2t + 3)*(t + 1) = 0
t = -1 (ignore)
t = 1.5 seconds
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