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When I was on holidays, I was told a story about how someone passing under a palmtree and almost got a coconut fall on his head.

Given that these palmtrees where about $20m$ high, we wondered at what speed the coconut was when it reached the ground... I took a pen and a paper and tried to remember what I knew about velocity, acceleration etc.

When I got home I found out that the formula to compute this is $v = \sqrt{2 * g * d}$, giving a final speed of about $71.3127 km/h$.

However, the question is not about the real formula, it is about my failed attempt at solving the problem. I tried reasoning about units and ended up with this:

  • I know that $g = 9.81 m/s^2$
  • I have a distance ($20 m$)
  • I want my result to be in $m/s$
  • Hum, hey look if I multiply $g$ by the distance I end up with $m^2/s^2$
  • And then I can simply do $\sqrt{m^2/s^2} = m/s$, giving the final formula for computing the speed to be $v = \sqrt{g * d}$

If I apply the math, I end up with a speed of $50.42 km/h$, which looked reasonable enough for me to believe I was right until I got home.

Ok, after all this introduction here comes the questions... I know that the reasoning is wrong, that I should integrate the speed etc and eventually end up with the correct formula, however I can't help thinking that what I'm computing has to have a meaning in the real physical world:

  1. What does the intermediate result in $m^2/s^2$ represent?
  2. What does the final result in $m/s$ represent?
  3. Should I avoid thinking about units when trying to solve a problem?
  4. Does every units manipulation result into something that means something in the real world, or does some manipulation just end up in non sense?
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Hi Silex. What you did is called dimensional analysis, and it's a strong tool in some senses. However, it doesn't provide the numerical constants, which happens to be $\sqrt{2}$ in this case. –  Ali Jul 28 '13 at 12:34
    
Unit manipulation results don't mean anything until you interpret them and correlate them with the context. For eg., Torque and Energy have the same units, but they represent two very different things. –  udiboy1209 Jul 28 '13 at 12:38

2 Answers 2

up vote 1 down vote accepted

1) What does the intermediate result in $m^2/s^2$ represent?

The gravitational acceleration times the distance fallen ($g \ d$) represents the kinetic energy per unit mass of the object.

2) What does the final result in m/s represent?

The quantity $g \ d$ equals $\frac{1}{2} v^2$ (see point 1) and therefore $\sqrt{g \ d} = v/\sqrt2$, about $0.7$ times the final velocity of the coconut.

3) Should I avoid thinking about units when trying to solve a problem?

No, please continue doing so! As others have remarked, you have discovered the power of dimensional analysis: with very little effort you can get an excellent insight into the functional dependencies. In your case you found out that the final velocity is some constant time the square root of $g$ times the distance.

4) Does every units manipulation result into something that means something in the real world, or does some manipulation just end up in non sense?

As long as your manipulations are right, there is always some meaning (although it can on occasions be a somewhat 'contorted' meaning).

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Good answer $\large{+1}$. –  Ali Jul 28 '13 at 14:32
    
Thanks! about point 2, does this $v / \sqrt{2}$ actually represents something visual or concrete? Because it's a square root of two, I can't help but visualize a square triangle having its hypothenuse being $71.31 km/h$ and the borders being $50.42 km/h$. Not sure what these borders would represent tho. –  Silex Jul 28 '13 at 17:25
    
@Silex - I think you are reading too much into this occurrence of $1/\sqrt2$. It is a direct consequence of the factor $1/2$ in the kinetic energy expression: physics.stackexchange.com/questions/27847/… –  Johannes Jul 28 '13 at 17:36
    
Alright, thanks for the link! –  Silex Jul 29 '13 at 8:28

This is a really long comment, which will address some of the questions.

What you did is basically what people call dimensional analysis. Now that you know the name, you can perhaps investigate more and learn more about it. Dimensional analysis can be a strong tool, it will give you the functional form of the final answer of your problem; without asking for too much heavy analysis(i.e. actually solving the problem). Since constant factors don't have units, dimensional analysis is unable to give them; you will need other means to derive them, e.g. special cases, experimental data or solving the problem completely. That constant factor is usually of $O(1)$ but in some rare cases it can be really big(or small).

David Morin gives a nice introduction about them here. He also has a lovely piece of poem:

“Your units are wrong!” cried the teacher.

“Your church weighs six joules — what a feature!

And the people inside

Are four hours wide,

And eight gauss away from the preacher!”

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Thanks, I'm happy to find out being pragmatic helps in physics too :) I would have liked to +1 your answer but unfortunately I don't have enough points at the moment. –  Silex Jul 29 '13 at 8:26

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