(adsbygoogle = window.adsbygoogle || []).push({}); 1. The hammer throw is a track-and0field event in which a 7.3-kg ball (the hammer), starting from rest, is whirled around in a circle several times and released. It than moves upward on the familiar curvinf path of projectile motion. In one throw, the hammer is given a speed of 29 m/s. For comparison, a .22 caliber bullet has a mass of 2.6g and, starting from rest exits the barrel of a gun with a speed of 410 m/s. Determine the work done to launch the motion of (a) the hammer and (b) the bullet.

2. KE=(1/2)mv^2

W= KE(final)-KE(initial)

3. I think I know my KE final, which would be just plugging in the numbers-

(1/2)*(7.3kg)*(29^2)=KE(final) of the Hammer... But How would I get my KE(initial) since I don't know what the initial velocity of the hammer is as its being whirled around in a circle.

Can someone help?

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# Homework Help: The Work-Energy Theorwm and Kinetic Energy

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