How Does the Law of Conservation of Energy Apply in Free Fall?

[LagrangeEuler]

Homework Statement

Particle goes to free fall from point $A$ to point $B$. $\upsilon_{A}=0$. How to define law of conservation of energy. Coordinate system is selected in such way that $y$ coordinate od point $A$ is zero. (See figure energy)

Homework Equations

For free fall $\upsilon_{B}=\sqrt{2gy}$.

The Attempt at a Solution

My problem is with formulation of law of conservation of energy. Energy of point $A$ is in this case $0$? Right? Because potential energy is zero ($y$ cordinate of point $A$ is zero) and in case of free fall $\upsilon_{A}=0$.
And energy of point $B$ is also zero $-mgy+m\frac{\upsilon^2}{2}=0$.
Am I right? I'm confused how the energy of the particle could be zero? Can you help me? Give me some other view?

 

[DrClaude]

I'm confused how the energy of the particle could be zero? Can you help me? Give me some other view?

What you take to be the zero of energy is arbitrary. At point B, the particle as $-mgy < 0$ potential energy (it has lost some potential energy) and $mv^2/2 > 0$ of kinetic energy (it has gained kinetic energy). (Note: I take it that the y-axis is pointing downwards.)

If you choose point B to be the zero of potential energy, then at point A, you have $-mgy + 0 > 0$ (since $y < 0$) and at point B $0 + mv^2/2$, with $mv^2/2 = -mgy$, which is exactly the same as before.

In both cases, you see potential energy converted into kinetic energy.