A construction worker tosses a hammer straight up to a co-worker who is working on a platform above him. The hammer leaves his hand at 7.26 m/sec, and travels 3.33 meters upward, at which point it is caught by the worker on the platform. (It is caught while still moving upward.)

Question

A construction worker tosses a hammer straight up to a co-worker who is working on a platform above him.  The hammer leaves his hand at 7.26 m/sec, and travels 3.33 meters upward, at which point it is caught by the worker on the platform.  (It is caught while still moving upward.)

a)	What is the acceleration of the hammer after leaving the first worker’s hand and before being caught by the second worker? i got a=0

b)	What is the velocity of the hammer at the instant that it is caught? i got v= 3.26

c)	How long did it take for the hammer to go from the thrower to the catcher? i got t=5

please help.

Steve · Answer

the acceleration is as usual: -9.8 m/s^2 due to gravity Considering h=0 when it was thrown, we have h(t) = 7.26t - 4.9t^2 = 3.3 I don't get your answers. In fact, I suspect a typo. What did you do?