1. The problem statement, all variables and given/known data A child stands on a balcony directly above a friend 5.00 m below. Each throws a ball toward the other with the same initial speed of 10.0 m/s. If each catches a ball at the same instant, which child threw a ball first? How much later was the second ball thrown? 2. Relevant equations I used x = xi + vi(t) + 1/2(a)(t^2) Where x is the vertical distance. 3. The attempt at a solution By just thinking about the question, the person below the balcony should throw the ball first if both people catch balls at the same instant. I'm not sure if vf can equal 0 m/s or not so I used x= xi + vi(t) + 1/2(a)(t^2). I let the positive x be upward. Bottom person: x = +5.00 m xi = 0 vi = +10 m/s a = -9.81 m/s^2 -4.905(t^2) + 10(t) - 5 Used the quadratic formula to get t = .87885 sec, 1.159878 Top person: x = 0.00 m xi = 5.00 m vi = -10 m/s a = +9.81 m/s^2 4.905(t^2) - 10(t) + 5 But doing this makes the t for top person and t for bottom person equal the same.