Question

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?

Answer 1

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).