When a parachute opens, the air exerts a large drag force on it. this upward force is initially greater than the weight of the sky diver and thus slows him down. suppose the weight of the sky diver is 819 n and the drag force has a magnitude of 1017 n. the mass of the sky diver is 83.6 kg. take upward to be the positive direction. what is his acceleration, including sign?

There's a lot of numbers here, which can tend to obscure our view of the important parts of the problem, and how to solve it.

The important parts of the problem are:  Mass = 83.6 kg, and Force = 1017 N.

The way to solve it is:  F = m a

That's Newton's 2nd law.  Divide each side by 'm' and it says  a = F/m .

acceleration = (force) / (mass)

acceleration = (1017 N) / (83.6 kg)

acceleration = 12.2 m/s²

If we say that UP is the POSITIVE direction, then the diver's velocity is negative (down), the force of air resistance is positive (up), and the acceleration is positive (up).

1= 26.17

2= 1.17

3=30.86

4=5.86

just took the assigment on edge