# Ball is thrown up at 88 ft/s from a height of 25 feet

###### Question:

ball is thrown up at 88 ft/s from a height of 25 feet. Path of ball is
y = -16t^2 + 88t +25. y is the height of ball above the ground t seconds after being thrown. when does it reach maxiumum height? (I got that at 2.75 s)
At what time does the ball hit the ground? (I set equation to =0, but I got 5.2 s and answer is 5.77 s)

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