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## Homework Statement

Problem 1.2 from the book

*Introduction to Quantum Mechanics (2e)*by Griffiths:

Suppose a rock is dropped off a cliff of height

*h*. As it

falls, a million photos are snapped, at random intervals. On each picture the distance the rock has fallen is measured.

a) Find the standard deviation of the distribution.

## The Attempt at a Solution

Starting with the equation of motion (assuming it falls in the positive x direction):

[tex]x(t)=x_0+v_0(t)+\frac{1}{2}at^2=\frac{1}{2}gt^2[/tex]

The time

*T*it takes to fall:

[tex]h=\frac{1}{2}gt^2[/tex]

[tex]T=\sqrt{\frac{2h}{g}}[/tex]

Probability a picture is taken in interval

*dt*:

[tex]\frac{dt}{T}=dt \sqrt{\frac{g}{2h}}=\frac{dx}{gt}\sqrt{\frac{g}{2h}}=\frac{1}{2h}\frac{1}{\sqrt{gt}}dx=\frac{1}{2\sqrt{hx}}dx[/tex]

So the probability density is:

[tex]\rho(x)=\frac{1}{2\sqrt{hx}}[/tex]

Expectation Value:

[tex]\mu=\int_0^h x \rho(x)dx=\int_0^h x \frac{1}{2\sqrt{hx}}dx=\frac{1}{2\sqrt{h}}\frac{2}{3}x^\frac{3}{2}=\frac{h}{3}[/tex]

Standard deviation:

[tex]\sigma^2=\int_0^h (x-\mu)^2 \rho(x)dx[/tex]

[tex]\sigma^2=\int_0^h (x-\frac{h}{3})^2 \frac{1}{2\sqrt{hx}}dx=\frac{4}{45}h^2[/tex]

[tex]\sigma=\frac{2h}{\sqrt{45}}[/tex]

Look right?