# How To Measure Amoung Of Energy Released

1. Sep 22, 2011

### phoenix_1326

To start with, I have never been in a physics class, nor am I an engineer. I am a computer programmer, and I love the science channel, so at best I am a bad amateur in this field of study. That being said...

I was thinking about calculations that could be used in games like Angry Birds, and I thought back to Einstein's E = MC2. I understand this formula is used to calculate the TOTAL energy released when mass is annihilated, but I thought there might be a way of using this formula to calculate the energy used when say a baseball is in flight. I had what I thought was a eureka moment when I came up with E = MC2 / A where A is the actual MPH the baseball is thrown at.

I looked up the mass of a baseball: 142 g. As I understand it, that makes M = .142
C = ~186,282
The pitcher can throw at 95 mph, so A = 95

As such...
E =.142(186,2822) / 95 = .142(34,700,983,524) / 95 = 4,927,539,660.408 / 95 = 51,868,838.53061053

E = ~5.27 Joules...or something like that

This of course is where I have run into a major issue. I have no idea if I am even looking at this problem the right way, or how it translates into even the digital world (let alone the real world where...according to other posts on this site...the "springy-ness" of the material comes into play as well as the size of the material vs. the size of the point of impact...ok, the head ache just hit).

Can someone break this down into a "general and laymen" understanding of what I'm looking at? I would greatly appreciate it...and I know my wife will as well...she knows I'm going to obsess on this for the next month or until I figure it out enough to satisfy me.

2. Sep 22, 2011

### ModusPwnd

This doesnt really make sense. Dividing an energy by a speed does not give you an energy.

3. Sep 22, 2011

### phoenix_1326

But speed is used to determine energy released in E = MC2. Why can't speed (which is a representation of a form of energy...kinetic?) be used? I'm not meaning to argue, I'm trying to understand...and I may be in WAY over my head at this point.

4. Sep 22, 2011

### Dr_Morbius

Modus is right, your calculation incorrect. The correct calculation to use is E=mv2/2. That gives you the kinetic energy based on the mass and the velocity of the object. Also, you need to use the metric system. You can't combine kilograms with MPH. You need to use kilograms and meters per second.

http://en.wikipedia.org/wiki/Kinetic_energy

5. Sep 22, 2011

### phoenix_1326

Ok, I think I understand now. I couldn't figure out what I was trying to measure earlier, so I just went with the first thing that popped into my mind. If I understood this correctly, then my earlier equation should look like this:

E=mv2/2

m = .142
v = 42.468799999999995

E = .142(42.4688^2) / 2 = .142(1803.59897344) / 2 = 256.11105422848 / 2 = 128.05552711424