# Question Based On Galileo's Law Of Falling Bodies

Galileo discovered that the distance fallen is proportional to the square of the time it has been falling.Why is it proportional to the square of the time and not just time? i.e $d \propto t^2$ why not just $d \propto t$

I know this question involves some common sense which i'am not able to use for some reason!

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I'm not sure what you mean by common sense here? If it was that obvious, we wouldn't be praising Galileo for discovering the law. It's also something contingent to a certain extent, i.e. it does not follow from some reasoning starting from "common sense" principles. – Raskolnikov Dec 25 '11 at 7:36
I tend to get shy when asking questions that "looks elementary" to professional physicists! – alok Dec 25 '11 at 7:48

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Don't be shy to ask. I'm just saying that someone had to do the experiment. You can't just reason out what nature is like. – Raskolnikov Dec 25 '11 at 9:55
@Raskolnikov: Indeed, the reason Galileo's law of falling bodies wasn't discovered two thousand years earlier was because no one bothered to do a simple experiment to test "common sense." There's a lesson to be learned here. – David H Nov 26 '12 at 1:20

## 2 Answers

It might be worth taking a look to the original text, Galileo's Discourses on Two New Sciences. The reasoning you're looking for is on the Third Day, a translation of which may be found online. The relevant parts are labelled Theorem I and Theorem II in the above-linked translation.

To derive that distance in a uniformly accelerated motion (e.g. free fall) goes as time squared, Galileo first argues that

The time in which any space is traversed by a body starting from rest and uniformly accelerated is equal to the time in which that same space would be traversed by the same body moving at a uniform speed whose value is the mean of the highest speed and the speed just before acceleration began.

This is argued on a graphical basis (see above link).

However, even though the pictures may look pretty similar to modern functional representations (e.g. of velocity vs. time) and arguments involve finding equal areas in different situations, the arguments never involve an actual calculation of an "area" with mixed units, which wasn't yet conceivable at the time (e.g. $m/s \cdot s = m$). In fact, the whole Third Day seems very convoluted precisely because the notion of velocity wasn't clearly numerical yet, since only commensurable (same unit) quantities could conceivably be operated with (added, divided,...). Proportions of non-commensurable quantities, however, could be compared (today we'd say they are dimensionless), as in

If a moving object traverses two distances in equal intervals of time, these distances will bear to each other the same ratio as the speeds

(earlier in the Third Day)

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Because the velocity increases with time, and the total time increases with time, so that the velocity times time increase as time-squared. It is dimensional analysis--- if you gain an increment of speed every time unit, you must multiply by the time twice to get a distance.

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Ah.Dimensional analysis.Right.But still having little bit of trouble understanding it. – alok Dec 25 '11 at 7:33
Maye it will help to know that the distance travelled is the area under the curve of the graph of the velocity vs. time, and this graph is a triangle for constant velocity? This is Galileo's argument from way back when. – Ron Maimon Dec 25 '11 at 14:11
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