1. The problem statement, all variables and given/known data A ball is thrown vertically upward with an initial speed of 11 m/s. One second later, a stone is thrown vertically upward with an initial speed of 25 m/s. (a) Find the time it takes the stone to catch up with the ball. (b) Find the velocities of the stone and the ball when they are at the same height. 2. Relevant equations 3. The attempt at a solution a = 9.8 Vi,ball= 11 m/s Vi,stone= 25m/s y= y0+ V0t+ 1/2 at2 y = 11t+ 1/2 (9.8) t2 y= 11t - 4.9 t2 ======= ball y = 25 (t+1) + (4.9 (t+1))2 y = 25t + 25 - 4.9 ( t2 + 2t + 1) y = 25 t + 25 - 4.9 t2-9.8t - 4.9 y= -4.9t2 + 15.2t + 20.1 ===== stone 11t - 4.9 t2 = -4.9t2 + 15.2t + 20.1 am i on the right track??