1. The problem statement, all variables and given/known data A stone of mass 3kg is thrown vertically in the air with an initial velocity of 2.8m/s. There are no forces acting on the stone except for that due to gravity and the thrower. Show that the overall time taken for it to return to its point of release is 4/7 seconds. 2. Relevant equations v = u + at v^2 = u^2 + 2as 3. The attempt at a solution Really all I want to know here is if I'm making more work for myself than is necessary. Taking the motion in two stages: 1. from the throwing until the stone is at momentary rest, 2. from the momentary rest until it reaches the thrower's hand again 1: u = 2.8, v = 0, a = -g t = (v - t)/a t = (0 - 2.8)/-9.8 t = 2/7 seconds. Now, at this point, I REALLY want to believe that I can just multiply that by two for its downward motion and that's it. But I'm in doubt: it's starting with zero velocity, not 2.8m/s velocity like in stage 1. Can I just assume that the time taken for it to return will just be the same again, 2/7 seconds, thus completing the task, or is there more to this?