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## Homework Statement

A given particle is confined to a certain potential. At a given instant that potential is turned off and the particle is accelerated by gravity. In the initial instant t = 0, when the potential is turned off, it has an initial speed of vz = 1 m/s.

In the instant t = 0 the particle goes through a slit with width y = 1 mm = 10^-3 m in the y axis. How much time does it take for the spatial uncertainty associated with y to double? Consider the law of propagation of uncertainty for the calculation of the spatial uncertainty of y at a given instant. Consider the particle a proton.

## Homework Equations

[;\sigma_{x}\sigma_{p} \geq \frac{\hbar}{2};]

[;\mathbf{p} = m\mathbf{v};]

[;v = v_0+at \,;]

[;\left |\Delta X\right |=\left |\frac{\partial f}{\partial A}\right |\cdot \left |\Delta A\right |+\left |\frac{\partial f}{\partial B}\right |\cdot \left |\Delta B\right |+\left |\frac{\partial f}{\partial C}\right |\cdot \left |\Delta C\right |+\cdots;]

## The Attempt at a Solution

I understood that, as the particle passes through the slit, we have [;\Delta y\right;] equal to the slit's width.

But, from that, I don't really get how to proceed, considering the particle is moving along the z axis. I can relate the errors along the z axis, but since there's no momentum along the y axis, I am kind of lost.

How am I supposed to use the propagation of uncertainty?

Thank you in advance. I don't expect to be just given the solution, I just want a little push forward. (I am new here, so please tell me if I'm doing anything wrong!)