1. The problem statement, all variables and given/known data A ball is thrown vertically upward with an initial speed of 20 m/s. Two seconds later, a stone is thrown vertically (from the same initial height as the ball) with an initial speed of 24 m/s. At what height above the release point will the ball and stone pass each other? 2. Relevant equations yf = yi + vit + ½ at2 3. The attempt at a solution The right solution : The ball is thrown 2 sec before the rock yball = ½ at2 + vit + yi yball = ½(-10)(t+2)2 + 20(t+2) yball = -5t2 + 20 yrock = ½ at2+ vit + yi yrock = ½(-10)t2 + 24t yrock = -5t2 + 24t yball = yrock (they pass each other) -5t2 + 20 = -5t2 + 24t t = 5/6 seconds yrock = -5(5/6)2 + 24(5/6) yrock = 17 meters but I don't know why he considered (t) of the ball (t+2) althought the ball is thrown 2 seconds before the stone??? isn't it true if : t of the ball = t-2 t of the stone = t ?