A 70Kg person jumps out of a plane and deploys a parachute, once the parachute is open the wind resistance acting on the person and the parachute is 900N. What is the direction and magnitude of the persons acceleration.

We start off by working out all the forces acting on the person and they're parachute, because they are joined together they can be considered as one object. We know the wind resistance and the only other force acting on them is the weight of the man (we assume the parachute is weightless). So the total weight is found using F = Ma. M is mass so is 70Kg and a is gravity so is 9.8m/s^2. Therefore the weight acting down is 686N. To find the acceleration we again use F = Ma but this time F is the resultant force acting on the person. So the resultant force is weight - wind resistance (we assume down is positive direction). Therefore the resultant force is 686 - 900 = -214. By rearranging the equation to get a = F/M the accleration can be worked out as -214/70 = -3.06m/s^2. And the negative sign indicates it is acting upwards i.e slowing the person down.