1. A parachustist jumps from an airplane and freely falls y=49.4 m before opening his parachute. Thereafter, he decelerates at a=2.02 m/s2. As he reaches the ground, his speed is 3.11 m/s. How long was the parachutist in the air? For this one I managed to get the solution already using a method someone else showed me. However, I fumbled a long time with this problem and got the wrong answers despite having a strategy in mind. What I did was the following. I split the problem up into two parts: before the parachute opens and after. These are the variables as I defined them at the beginning of the problem: v0=0, s0 = 0, for when parachutist jumps out plane. v1 = ?, s1 =49.4m, a1 = 9.8 m/s2, t1 = ?, upon parachute release v2 = 3.11 m/s, s2 = ?, a2 = -2.02 m/s2, t2 = ?, upon reaching ground 1. First I tried to find the time t2 between parachute release and reaching ground. 0.5at2 + v2t + s1 0.5(-2.02 m/s2)t2+(3.11 m/s)t +49.4m using the Q-formula I got t2=8.71 2. Then I tried finding v1 v2=v1+at2 3.11 m/s = v1 + (-2.02 m/s2)(8.71s) v1 = 20.7 m/s 3. Using v1, I could then find t1 t1 = (v1-v0)/a t = 2.11s The red part I think is where the mistake was. For that formula, I had to use v1, which I didn't know, but even if I did, would this approach still work? Plugging in values for v1, which I found later using a different method, still gave me wrong figures for t. Rather than telling me how I could obtain a solution, please show me the mistake with this strategy and whether I could have tweaked it to make it work. I have struggled a long time with kinematics concepts and I admit my problem solving skills are subpar, so a physical description of what is occuring would be very helpful. Thanks!