Looking for a simplified explanation for why you cannot measure velocity and pin point location

Here is a way to see this relationship from a physical standpoint.

Imagine we want to pinpoint the position of an electron. To do so, we shoot another electron at it and see how it bounces off. By back-tracking the trajectories of the two particles after the collision, we can determine where the electron was at that moment, but note the following: because the collision caused it to move, the electron isn't there any more. What to do?

Well, we could lob our electron bullet more gently at the target, so the target will move less when struck. But slowing down the electron bullet means reducing its energy, and that means increasing its equivalent wavelength. And that means the bullet becomes a less precise tool for pinpointing the location of the target. What to do?

Well, we could increase the velocity of the electron bullet, which shortens its wavelength and increases its precision, but then it will hit the target harder and send it out of position with a significant velocity.

This means that anything we do do more precisely locate the electron not only affects its location but also affects its velocity, and anything we do to affect its position and its velocity less reduces the accuracy of our position measurement.

For very very tiny things like electrons, we are stuck with this fundamental tradeoff and we can't do anything about it. Luckily, the bigger the object becomes, the less important this effect is and so for things like baseballs and cars we do not have to worry about it at all.

There are a lot of possible answers to that question, but one can answer it just from the mathematical point of view.

Take a function $fin L^2(mathbb{R}^n)$ and using the following definition of the Fourier transformation from the position to the momentum space

$$ hat{f}(p)=frac{1}{(2pi)^{frac{n}{2}}}int_{mathbb{R}^n}e^{-ipcdot x} f(x); mathrm{d}^nx $$

For $||f||_2=1$ one can derive that: $$ left(int_{mathbb{R}^n}x^2|f(x)|^2;mathrm{d}^nx right)left(int_{mathbb{R}^n}p^2|hat{f}(p)|^2;mathrm{d}^nxright)geqfrac{1}{4} $$

Then we can notice that those integrals could represent the square of the variance of the expectation value of momentum and position. Because of that equation (ii) can be written as $(Delta x)^2 (Delta p)^2geqfrac{1}{4}$. If we now take the square root of that and set $hbar=1$ we get the famous Heisenberg uncertainty principle: $$ Delta x Delta p geq frac{1}{2} quad text{for} quad hbar=1 $$ An other way would be for example to argument that on quantum mechanical scale one can not measure position or velocity/momentum without changing the quantum state of the system, making it impossible to measure both because after measuring one the state of the system has changed. But I like the idea of a mathematical concept.