Linear velocity of arm segments in volleyball spike

[stuart_5]

Below is a question I will need to know well for my exam coming up. I don't even know where to start or what the steps are? I am very overwhelmed here...

A volleyball spike begins with the arm overhead, the shoulder and elbow are flexed and the wrist is hyperextended. The upper arm is .6m and the forearm is .3m and the hand is .1m long. The time of spike is .1 seconds, and the changes in position of each joint is: Shoulder = 30 degrees, Elbow = 70 degrees, Wrist = 140 degress.
What is the linear velocity at the end of the distal endpoint of each segment at the end of the segment's rotation?

Any assistance would be greatly appreciated... Thank you.

 

[dicerandom]

I think you're supposed to consider the arm-hand system as being three separate rigid rotators which are connected end to end. I also think that you're supposed to assume they do not undergo angular acceleration. So, in the space of 0.1 seconds the shoulder rotates 30 degrees, on the end of that the elbow rotates 70 degrees, and then the wrist rotates 140 degrees. For each rotation, working from the shoulder out, you need to figure out what the linear velocity at the end of the relevant arm segment is, then add them all up to find the final velocity of the volleyball.