Answer to Question #5153 in Mechanics | Relativity for Adam
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?
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).