A ball is thrown vertically upward with an initial velocity of 96 feet per second.The distance s (in feet) of the ball from the ground after t second is s(t)=96t-16t².at what time t is the ball more than 128 feet above the ground?

Answer 1

-16t² + 96t - 128 = 0 ... t² - 6t + 8 = 0

solve for t ... the two times are upward and downward
...the ball is above 128 ft between the times