# Physics

Bob starts spinning his hammer at a rate of 2.00 rad/s^2. At the same time, he starts running towards Jim at 4.00 m/s^2. The distance from the end of the hammer to Bob's elbow is 1.20 m. If Bob takes 3 seconds before he launches the hammer at Jim, what is the resultant velocity of the hammer when it leaves Bob's hand?

My attempt.

tangential acceleration = (1.20m)(2.00rad/s^2) = 2.4 m/s^2
resultant/total acceleration = 2.4 m/s^2 + 4.00 m/s^2 = 6.4 m/s^2
change in velocity = (6.4 m/s^2)(3.00s) = 19.2 m/s

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