What is the average of fall distances

[Ilikechiken]

[no template, as this post was moved from here from the Quantum Mechanics forum]

In griffiths 2nd quantum mechnics,

problem : Suppose I drop a rock off a cliff of height h. As if falls, I snap a million photographs, at random intervals. On each picture I measure the distance the rock has fallen.

Question : What is the average of all these distances? That is to say, what is the time average of the distance traveled?

I know p(x) is 1/2(hX)^(1/2) and can solve the problem like : Integral(from 0 to h)x dx/2(hX)^(1/2)

other method is : Integral(from 0 to T) x dt/T

but I don't know how it's possible. that means p(t) = 1/T .. but T is constant{ (2h/g)^1/2}

and I don't know meaning of dt/T

 

[HallsofIvy]

Here's how I would do that- the constant acceleration due to gravity is -g. The velocity, at time t isv(t)= -gt+ v0. Since the rock was "dropped", rather than thrown up or down, v0= 0 and v(t)= -gt. Then the distance traveled in t seconds is $s(t)= -(g/2)t^2+ h$. The rock will hit the bottom when $s(t)= -(g/2)t^2+ h= 0$ so when $t= \sqrt{2h/g}$. The average of a function, f(t), for t between a and b, is $\int f(t)dt/(b- a)$ so the average height, for t between 0 and $\sqrt{2h/g}$ is $$\frac{\int_0^{\sqrt{2h/g}}( -(g/2)t^2+ h) dt}{\sqrt{2h/g}}$$

 

[mfb]

I don't understand your notation - I think you mix different things here. p(t) in the first approach looks like the position, while in the second approach it seems to be a probability to take a picture at a specific time. But then the integral should have the position as function of time, not x.

 

[Crush1986]

I think the problem and his approach was worded so strangely. I had to just close the book and think about it by myself, and figure out how I'd define each piece of the problem mathematically. Then how I could manipulate each piece in order to integrate it. Turned out it was pretty much identical.

Sorry I necro'd this thread. I just had to see if anyone else struggled with this example. Then I wanted to say how I approached the problem so anyone else in the future can maybe try the same type of thing if they were also confused.