SOLUTION: A baseball player throws a ball from the outfield toward home plate. The ball's height above the ground is modeled by the equation y =-16x^2+48x+6, where y represents height, in f

Question 313975: A baseball player throws a ball from the outfield toward home plate. The ball's height above the ground is modeled by the equation y =-16x^2+48x+6, where y represents height, in feet, and x represents time, in seconds. The ball is initially thrown from a height of 6 feet. how many seconds after the ball is thrown will it again be 6 feet above the ground? what is the maximum height, in feet, that the ball reaches?
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!

You can do this one in your head. y will equal 6 whenever

. Obviously the first time this occurs is at time zero, just when the ball is thrown. The next time is 48 divided by 16 or 3 seconds. By symmetry, the maximum height is reached halfway between 0 and when the ball returns to 6 feet, namely

seconds. The height will be -36 + 96 = 60 feet.

John

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