A .72 kg arrow is pulled back in a bow and pointed straight up into the air. The arrow has 162 J of elastic potential energy. How fast will the arrow be moving when it is released from the bow?
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Expert's answer
2011-11-17T12:00:54-0500
According to the energy conservation law the potential elastic energy is transformed into the kinetic energy: Ep = (m*V^2)/2. So, V = sqrt(2*Ep/m) = sqrt(2*162/0.72) = 21.2 (m/sec).
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