1. The problem statement, all variables and given/known data A referee throws a ball in a basketball game. When the ball was in his hand the heigh was 1,6 meters above the ground. The ball is caught 0.8 seconds later by a jumping player. The ball is 2.7 meters above the ground and on the way down. what was the initial velocity when the ball left the referee's hand? 2. Relevant equations s=v0t + at^2/2 v=v0 + at 3. The attempt at a solution I am guessing that the basketball player caught the ball when it was at its highest point and therefor the velocity was 0 m/s. 0 = v0 + at at = - V0 -9,86 * 0,8 = - v= 9,82 * 0,8 = v this answer is wrong so i tried again with the second equation and that gave the right anwer. Why is that?