Free fall with a partial time provided

[ConfusedStudentx10E9]

Homework Statement

A tree has two devices attached to its trunk which are triggered when objects pass in front of them. The devices are vertically aligned and are placed 1.50 m apart, with the bottom one being 0.65 m above the ground. You climb the tree and decide to experiment to find out how high h0 above the ground you are. You release your wallet from your level so that it will pass in front of the motion detection devices. Once you've climbed down, you read that the lower device was triggered 0.194 s after the first one. Neglecting air friction and assuming that the wallet did not hit any branches on its way down, how high h0 had you climbed the tree?

Relevant Equations

h=1/2gt^2

I'm stuck on this problem, I've tried to follow techniques for similar questions, namely I seem to be struggling with these questions where I have to use an equation inside an equation. I've attached photos of my process so far, but obviously, I'm not getting the right answer because what I'm trying to solve for is being canceled out of the equation unless I messed up somewhere in rearranging the equations.

Help is much appreciated.

 

[kuruman]

You know that the wallet takes 0.194 s to travel 1.50 m between sensors. You can write an equation relating the two quantities which will have one unknown. What quantity is that? How can it help you solve the problem if you know its numerical value?

 

[ConfusedStudentx10E9]

You know that the wallet takes 0.194 s to travel 1.50 m between sensors. You can write an equation relating the two quantities which will have one unknown. What quantity is that? How can it help you solve the problem if you know its numerical value?

is it the y=y0+vot-1/2gt^2 equation?

I'm confused because v0 would be zero since it was dropped, and I already know y values for this specific time, so everything in this equation is already known right?

 

[kuruman]

I am talking about the transit time $T = 0.194~$s required for the wallet to travel distance $L=1.50~$m. What equation can you write that relates $T$ and $L$?

 

[ConfusedStudentx10E9]

I am talking about the transit time $T = 0.194~$s required for the wallet to travel distance $L=1.50~$m. What equation can you write that relates $T$ and $L$?

We haven't used any equations with T or L in this course - unless they have called it something different, but I don't see anything on my eqaution list that would compare.

 

[kuruman]

In my notation I use $T$ for 0.194 s which is the time it takes the wallet to pass between sensors and $L$ for the 1.50 M that is the distance between sensors. You can use whatever symbol you are familiar with. This is algebra.

My point remains, what equation can you write that relates the time it takes the wallet to pass between sensors and the distance between sensors, whatever symbol you choose to attach to them?

 

[ConfusedStudentx10E9]

In my notation I use $T$ for 0.194 s which is the time it takes the wallet to pass between sensors and $L$ for the 1.50 M that is the distance between sensors. You can use whatever symbol you are familiar with. This is algebra.

My point remains, what equation can you write that relates the time it takes the wallet to pass between sensors and the distance between sensors, whatever symbol you choose to attach to them?

the only one i have given on the equation sheet is y=y0+vot-1/2gt^2

or v^2=v0^2-2g(y-y0)

I don't have anything else that includes a Hight distance.

 

[kuruman]

Use the first one but adapt it to the motion between sensors.

 

[ConfusedStudentx10E9]

Use the first one but adapt it to the motion between sensors.

so 1.5=-1/2gt^2
?
but now I'm not solving for anything becaaue I already know all these

 

[jbriggs444]

so 1.5=-1/2gt^2
?
but now I'm not solving for anything becaaue I already know all these

If you know the time taken to traverse a known distance, that gives you a velocity. An average velocity.

Does that make sense? If so, we can proceed.

 

[ConfusedStudentx10E9]

If you know the time taken to traverse a known distance, that gives you a velocity. An average velocity.

Does that make sense? If so, we can proceed.

yes v=d/t

 

[jbriggs444]

yes v=d/t

OK, good. So we have an average velocity over some range.

Now we need to be more specific. It is an average over what time range? I am not fishing for a numeric answer. I am fishing for an answer more like "the time from when <some mumble> to the time when <some other mumble>".

[I gave you that is an average over time for free].

 

[ConfusedStudentx10E9]

OK, good. So we have an average velocity over some range.

Now we need to be more specific. It is an average over what time range? I am not fishing for a numeric answer. I am fishing for an answer more like "the time from when <some mumble> to the time when <some other mumble>".

the time from when it passed the first sensor to the time when it passed the second sensor.

v^2=v0^2-2g(y-y0)

where V0 is zero, because it was dropped, and y= 0.65 (the bottom sensor) and y0= 2.15 (the top sensor)

This?

 

[jbriggs444]

the time from when it passed the first sensor to the time when it passed the second sensor.

v^2=v0^2-2g(y-y0)

where V0 is zero, because it was dropped, and y= 0.65 (the bottom sensor) and y0= 2.15 (the top sensor)

This?

Stop with the squared velocities already. You do not need them. This is way easier than that. You have answered correctly: The average velocity is from the time when it passed the first sensor to the time when it passed the second sensor.

Where do we stand now? We know an average velocity. It would be more helpful (and would pretty much immediately solve the problem) if we knew an instantaneous velocity at some event. So we need to work toward such an answer.

There are a couple of ways to proceed. I have one in mind.

We are under constant acceleration, yes?
We also know the acceleration rate, yes?

Next question:

Given constant acceleration, is the average velocity over a time interval equal to the instantaneous velocity at the midpoint of that time interval?

Alternately:

Given constant acceleration, is the average velocity over a time interval equal to the mean of the velocity at the beginning of that interval and the velocity at the end?The general approach that I am taking here is a meet-in-the-middle tree search. We are trying to work forward from what we know to what we can calculate. We are trying to work backward from what the problem asks for to a set of things sufficient to calculate that answer. If we can meet in the middle where we know a sufficient set of things then we can actually do the calculations and solve the problem.

At this point, we know that we can calculate an average velocity over the time interval for the wallet to pass the two sensors. We know (or I confidently assert anyway) that if we knew the velocity of the wallet when it passes one of the sensors that we could determine the height above that sensor from which it was dropped. [At which point you will indeed be squaring a velocity -- just not an average velocity]

 

[PeroK]

the only one i have given on the equation sheet is y=y0+vot-1/2gt^2

or v^2=v0^2-2g(y-y0)

I don't have anything else that includes a Hight distance.

This is why it's important to know some mathematics. You have your basic equation:$$y = y_0 - \frac 1 2 gt^2$$And you have $y$ at two different times$$y_1 = y_0- \frac 1 2 gt_1^2$$$$y_2 = y_0 - \frac 1 2 gt_2^2$$Whenever you have two equations you can try adding or subtracting them. If we subtract the second equation from the first we get:$$y_1 - y_2 = y_0 - \frac 1 2g t_1^2 - y_0 + \frac 1 2 gt_2^2 = \frac 1 2 g(t_2^2 - t_1^2)$$Now, one thing you must know and never forget is that, for any numbers, we have$$a^2 - b^2 = (a-b)(a+b)$$That's one of the most useful things in all of mathematics. If we apply it here we get$$y_1 - y_2 = \frac 1 2 g(t_2 - t_1)(t_2 + t_1)$$Now, in this case, we know $y_1 - y_2$ and $t_2 - t_1$, so we can calculate $t_2 + t_1$. And, if we know $t_2 - t_1$ and $t_2 + t_1$, then we can solve those simultaneous equations to calculate both $t_1$ and $t_2$. Then we can get $y_0$ directly from either equation.

There may be quicker ways to get the answer, but that shows how basic mathematical technique can crack these problems. And why just plugging numbers into equations may not get you very far.