I seem to have some irregularities with a specific problem which goes like this: "A stone is thrown upward from the edge of a cliff, with a speed of 70m/s. If, when it is 120m above the hand on its return journey, a second stone is dropped over the edge of the cliff, when and where will the stones be together?" A nice little illustration: http://img381.imageshack.us/img381/8154/testing6dv.th.jpg [Broken] A graph to illustrate the velocities of the 2 stones: http://img403.imageshack.us/img403/3606/testing6cg.th.jpg [Broken] Now first, what I did was calculate the final velocity of stone #1: 2a + Δd = Vf^2 - Vi^2 2(10m/s^2) + (Δd) = Vf^2 - (-70m/s)^2 Vf^2 = 2500 Vf = 50m/s Using this final velocity, I found the distance of stone #2: 2a + Δd = Vf^2 - Vi^2 2(10m/s^2) + (Δd) = (50m/s)^2 - (0m/s)^2 Δd = 2480m Now, the only time in which the stones can be together is when their final velocities are the same. Knowing this, we use this equation: Δd = (vi)(Δt)+1/2a(Δt)^2 (2480m + 120m) = (50m/s)(Δt) + 1/2(10m/s)(Δt)^2 (In this step, the total distance is the distance from stone 2 but also added from stone 1) 52/5 = Δt^3 Δt = 2.2 The time in which the two stones meet is 2.2s but the answer is 2.4. Did I do something wrong in my process, or is my approach completely wrong?