Physics paper, picture analysis

In summary: The horizontal distance is also measured between the two consecutive images. The average speed of the ball can then be calculated using the formula:v_{avg}=\frac{x}{t}where x is the horizontal distance and t is the time interval found earlier. This average speed represents the flash rate, as requested.In summary, the conversation revolves around answering questions about a picture to be used in a final paper for physics. The main focus is on determining the flash rate and the initial velocity of the ball in the picture. The approach involves using kinematics equations and a scaling factor to estimate the distances and times. The final answer for the flash rate is approximately 0.043 seconds between each image, and the initial velocity is approximately
  • #1
mikky05v
53
0

Homework Statement


This isn't really a homework probelm, I'm trying to answer some questions about a picture to use in my final paper for physics. I thought it would be a fun thing to have thrown in there but I can't seem to figure out how to start.
http://img1.photographersdirect.com/img/8998/wm/pd1304871.jpg [Broken]
The questions I chose to answer were What is the flashrate and how fast is it moving when it leaves the man's hand.



Homework Equations


I don't know what equations i need but I'm sure it would be some of the kinematics equations. I do know that at the top of each bounce it's x and y movement = 0 or something like that. And i know i need to come up with a scaling factor which i will put up in about 5 minutes when my printer finishes printing out this picture. Any advice or whatever for other things that could help would be awesome. I am not a physics major lol.


The Attempt at a Solution

 
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  • #2
From the thickness of the fingers one can estimate the VERTICAL distance the ball falls from one picture of the ball to the next one.
 
  • #3
grzz said:
From the thickness of the fingers one can estimate the VERTICAL distance the ball falls from one picture of the ball to the next one.

that's exactly what i was about to try but i had to print a ruler.. turns out the only one i have is for architecture and is useless lol
 
  • #4
I suppose that the ball is released from rest.
 
  • #5
grzz said:
I suppose that the ball is released from rest.

it might be but if it was it should drop straight down not on a slight angle and the pictures would be closer together at the beginning like they are at the tops of the bounces
 
  • #6
from my measurements the width of the finger in the picture is .5cm exactly and the average real finger width is 2cm. which gives a scalinging factor of real/picture= 4
the ball is 9/10cm so multiplied by the caling factor it is 3.6cm. the distance between the first ball an the second is 3cm*sf=12cm
I took the distance of the 2nd bounce (because it has a clear picture of the top of the bounce where the vertical=0) and found the the ball moves 21.8cm (thats picture*sf) vertically

Now what?
I think i need a kinematics equation of some sort.. I can determine that the change in time is equal to sqrt(2(change-y)/gravity) but i don't know hwat forumal that is derived from it's just something i have written down. Can someone tell me what formula this is from? it gives me a time between each ball of approx .21seconds.. so that answers my first question of flash rate.

How do i find the initial velocity as it leaves his hand
t=.21seconds displacement of y= 12cm displacement of x=3.2cm
 
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  • #7
I have an answer.
I used the formulas:
x=Vix*t+.5g*t^2
=>Vix=(x-.5gt^2)/t
and Viy=(y-.5gt^2)/t

Since I measured my x to 3.2cm I got 55.61cm/s
y was 12cm so 265.12cm/s

Then I used pythagorean's theorem |V|= sqrt(x^2+y^2) = 270.90 cm/s

can anyone give me some feedback?

Correction: t= .042 not .21 bc you have to divide .21 by the 5 images in that span. The above numbers have been adjusted and should now be correct.
 
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  • #8
mikky05v said:
it might be but if it was it should drop straight down not on a slight angle ...

You are right.

Allow me to correct my previous post.

The ball CANNOT fall from rest since it must have a component of velocity in the horizontal.
 
  • #9
The flash rate can be found from the first and second position of the image of the ball.

The formula

y = v[itex]_{i}[/itex]t + [itex]\frac{1}{2}[/itex]gt[itex]^{2}[/itex]

is used in the VERTICAL direction.

The vertical direction is chosen since in this direction we know the initial speed v[itex]_{i}[/itex] which is assumed to be 0 m/s.

The value of y is estimated from the picture and the value of g is known. Hence the time interval of the two consecutive images can be found.
 

1. What is a physics paper, picture analysis?

A physics paper, picture analysis is a scientific method of examining and interpreting an image or diagram related to a physics concept or experiment. It involves using knowledge of physics principles and equations to analyze the image and draw conclusions about the physical phenomena depicted.

2. How is a physics paper, picture analysis different from a regular picture analysis?

A regular picture analysis may focus on the aesthetic or artistic elements of an image, while a physics paper, picture analysis specifically applies the principles of physics to understand and explain the scientific content of the image.

3. What are the benefits of performing a physics paper, picture analysis?

Performing a physics paper, picture analysis can enhance understanding of complex physics concepts, improve critical thinking and problem-solving skills, and provide visual representations of physical phenomena that are difficult to imagine or visualize.

4. What are some common techniques used in a physics paper, picture analysis?

Some common techniques used in a physics paper, picture analysis include identifying and labeling key components of the image, applying equations and principles to calculate values or make predictions, and comparing the image to real-world observations or experiments.

5. Can anyone perform a physics paper, picture analysis?

While a basic understanding of physics principles is helpful, anyone with an interest in physics can perform a physics paper, picture analysis. However, it is important to have access to accurate information and resources in order to effectively analyze the image and draw accurate conclusions.

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