# Velocity and transfer of energy for a body in free fall

I just need to check if my understanding of the transfer of gravitational potential energy to kinetic energy is correct. Is anything wrong below?

Say if a body were suspended at a height h above the surface. It has potential energy $E_P=mgh$. When the body is released from this height, the potential energy is transferred into kinetic energy $E_K=\frac12 mv^2$.

$$mgh = \frac12mv^2$$ $$gh = \frac12v^2$$ $$2gh = v^2$$ $$v = \sqrt{2gh}$$

Is this saying for any object (ignoring the effects of drag) its velocity is the same?

Factoring in drag, we know that drag $\propto$ velocity$^2$. So if the object (falling from the same height) has a greater surface area/size, its drag will be greater and velocity will be less.

Yes. Gravity is not a force in the Newtonian sense. The potential energy of the body isn't stored in the gravitational field or the Earth, it's in the body itself, in its mass-energy. When you drop the body the potential energy gets converted into kinetic energy, and the mass reduces. When you radiate the kinetic energy away, you're left with a mass deficit, see Wikipedia. So $E_P=mgh$ and $E_K=\frac12 mv^2$ are a little misleading.