Answer:
The order = [tex]\mathbf{\sqrt{\dfrac{h}{2 \pi m}} \sqrt[4]{\dfrac{H}{g}} \sqrt[4]{\dfrac{1}{2}} }}[/tex]
Explanation:
To miss the crack at a given distance is apparently not the same as the uncertainty that occurred in the distance while falling from the tower. However, it is believed that the uncertainties in both cases appear to be the same.
So, let's work it out together
According to Heisenberg's uncertainty principle:
[tex]\Delta s. \Delta p =\dfrac{h}{2} =\dfrac{h}{4 \pi}[/tex]
Also; if we recall from the equation of motion that:
[tex]v = u + at ---(1) \\ \\ v^2 - u^2 = 2as --- (2) \\ \\ s = ut + \dfrac{1}{2}at^2 --- (3)[/tex]
So, if u = 0 and a = g
Then;
[tex]v = gt --- (1) \\ \\ v^2 = 2gs - - - ( 2) \\ \\ s = \dfrac{1}{2}gt^2 --- (3)[/tex]
From (2)
Making (s) the subject, we have:
[tex]s = \dfrac{v^2}{2g}[/tex]
[tex]s = \dfrac{p^2}{2gm^2}[/tex]
By differentiation;
[tex]ds = d (\dfrac{p^2}{2gm^2})[/tex]
[tex]ds = \dfrac{2pdp}{2gm^2}[/tex]
[tex]\Delta \ s = \dfrac{p \Delta p}{gm^2 }[/tex]
where;
[tex]\Delta p = \dfrac{h}{4 \pi \Delta \ s}[/tex] from uncertainty principle
This implies that:
[tex]\Delta s = \dfrac{p(\dfrac{h}{4 \pi \Delta s }) }{gm^2}[/tex]
[tex]\Delta s = p(\dfrac{h}{4 \pi gm^2 }) \times \dfrac{1}{ \Delta s}}[/tex]
[tex](\Delta s)^2 = \dfrac{hmv} {4 \pi gm^2 }[/tex]
here;
v = 2gH
So;
[tex](\Delta s)^2 = \dfrac {h \sqrt{2gH} }{4 \pi gm }[/tex]
[tex]\mathbf{(\Delta s)^2 = \sqrt{\dfrac{h}{2 \pi m}} \sqrt[4]{\dfrac{H}{g}} \sqrt[4]{\dfrac{1}{2}} }[/tex]
Thus, the order = [tex]\mathbf{\sqrt{\dfrac{h}{2 \pi m}} \sqrt[4]{\dfrac{H}{g}} \sqrt[4]{\dfrac{1}{2}} }}[/tex]
