When a body is dropped from a certain height and loses an amount UUU of its potential energy, it acquires a velocity vvv. Using the conservation of mechanical energy, the mass mmm of the body can be expressed as:

m=2Uv2m=\frac{2U}{v^2}m=v22U​

This relation comes from equating the lost potential energy to the kinetic energy gained by the body, where the kinetic energy is 12mv2\frac{1}{2}mv^221​mv2 and the loss in potential energy U=mghU=mghU=mgh (height difference). By rearranging, the above formula is obtained.

Explanation

• Initially, the body has potential energy mghmghmgh.
• After dropping and losing an amount UUU of this potential energy, the body converts this energy into kinetic energy.
• The kinetic energy at that moment is 12mv2=U\frac{1}{2}mv^2=U21​mv2=U.
• Solving for mass mmm gives m=2Uv2m=\frac{2U}{v^2}m=v22U​.

This formula assumes no other energy losses (e.g., air resistance) and purely gravitational energy conversion during the fall.