1. The problem statement, all variables and given/known data A sky diver and his parachute weigh 890N. He is falling vertically at 30.5m/s when his parachute opens. With the parachute open, the magnitude of the drag force (in Newton) is 0.5v2. What is the magnitude of the sky diver's acceleration at the moment the parachute opens? 2. Relevant equations F=ma 3. The attempt at a solution So I figured the drag force is acting up and =0.5v2, weight acts down and = 890N, the sky diver accelerates downwards at (W/g)a So I got to an equation: 0.5v2-890=-(890/9.81)a I just tried plugging in the velocity at which he's falling (because at the instant the parachute opens, the velocity doesn't change yet) and got a=4.89. But the answer is supposed to be 24g. I don't really get that, at all. Help?