A man on the edge of a cliff H = 42 m high throws a ball directly upward. It returns past him 2 s later. (H actually is the height of the point of release of the ball above the base of the cliff. Neglect air resistance.) (Hint - Gravity causes a downward acceleration at the rate g = 9.81 m/s2.) There are multiple parts, but I'm only stuck on this first part. With what initial speed did the man throw the ball? So at first, I thought it was zero, but computer told me it was wrong so its not. For this problem, I'm taking it to be upward is positive in the y direction, and downward past the man is negative. I'm going to use the equation x-x0=v0(t)+.5(a)t^2 where x-x0=0 t=2sec a=9.8m/sec^2 and I get v0=-9.8. The correct answer is 9.8. So what gives?