Physics paper, picture analysis

[mikky05v]

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
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

 

[grzz]

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.

 

[mikky05v]

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

 

[grzz]

I suppose that the ball is released from rest.

 

[mikky05v]

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

 

[mikky05v]

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

 

[mikky05v]

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.

 

[grzz]

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.

 

[grzz]

The flash rate can be found from the first and second position of the image of the ball.

The formula

y = v_{i}t + \frac{1}{2}gt^{2}

is used in the VERTICAL direction.

The vertical direction is chosen since in this direction we know the initial speed v_{i} 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.