A wave whose wavelength is fixed to exactly one value goes on in the same shape without end. Having neither a beginning nor an end, there is no answering where it is.
For a matter wave, having a fixed wavelength is the same as having a fixed momentum, since p = \dfrac{h}{\lambda}. A particle whose momentum is exactly fixed has no fixed position at all.
Reading where it is likely to be from |\psi|^2, this wave has |\psi|^2 at the same height everywhere. Every place is equally possible.
Superpose many waves of slightly different wavelengths and in one place the crests line up and grow large, while further off they cancel and vanish. A wave gathered into one lump like this is called a wave packet.
The more waves are superposed, the narrower the lump becomes. That is, the better fixed the place is.
But raising the number of waves also widens the range of wavelengths it holds. The more the position is narrowed, the more the momentum spreads. That is what uncertainty amounts to.
Send an electron through a narrow slit. Writing the width of the slit as \Delta x, its position at the moment it passes is fixed to within that width.
The wave leaving the slit diffracts and spreads. The narrower the width, the larger the angle of the spread. Having spread means the direction of travel varies, which means the sideways momentum varies.
The angle at which the first dark line of the diffraction appears is \sin\theta = \dfrac{\lambda}{\Delta x}. The spread in sideways momentum makes \Delta p about p\sin\theta = \dfrac{h}{\Delta x}, and multiplying the two makes \Delta x \Delta p about h.
Defined properly and worked out, this comes to \Delta x \Delta p \geqq \dfrac{h}{4\pi}. However small one of them is made, the other grows, and the product does not fall below this value.
This is not a matter of measuring badly. Nor is it something a better instrument would settle for both at once. What the formula says is that fixed values for both are not held at the same time.
A pair of \Delta x and \Delta p can only be placed above this curve. The wave packet that lies exactly on the boundary is the one with the smallest product.
\dfrac{h}{4\pi} is a number as small as 5.3 \times 10^{-35}. Since \Delta p = m\Delta v, the larger the mass, the smaller the lower bound on \Delta v.
Fix the position of a 150 g ball to a precision of 0.1 mm and the lower bound on the spread in speed is about 10^{-30} m per second. That is not an amount anything can measure.
Do the same for an electron and the story changes. Confine it to the size of an atom, 10^{-10} m, and the spread in speed comes to about 10^{6} m per second. At that size one cannot say that an electron inside an atom goes around at a fixed speed.