# Homework Help: A simple uncertainty question about height.

1. Sep 13, 2014

### psstudent

So this is a physics question, that im trying to solve. The premise seems simply but im getting caught up on the uncertainty.

"Something from the top of a platform takes (2.0 ± 0.1)s to fall to the floor, assuming the acceleration due to gravity is 10 m/s2, what is the height."
Now the way I went about this was I used displacement for distance as they should be the same ( i think) because the object has changed position. So i used the formula S= Ut + .5at2 where s is displacement/distance.

Now assuming this is the right formula to use ( please tell me if this is incorrect) then I figure that i can use the maximum and minimum values of the time which would be 2.1 and 1.9 seconds respectively to calculate the height for what could be the highest and lowest values.
so my working goes like this:

S= ut + .5at2
= disregarding the ut as the initial im assuming would be zero and thus make the part of the formula before the plus sign negligible

thus = 5m/s2 X 2.12 and 5m/s2 X 1.92

those values gave me 22.05 and 18.05 as the highest and lowest heights so i averaged them together and used the magnitude between the average and each height to find the plus minus and ended up with 22.05 + 18.05 =40.1/2 = (20.05 ± 2)m

Now that made sense to me however the uncertainty seems to large to me as 2 meters is a big difference when if my answer the height is only 20.05 so thats almost a 10% uncertainty.

My friend also tried to solve this (he is in the same class). He used the same formula but instead of doing it like me (finding both the maximum and minimum value) he just used the formula and made the uncertainty a part of the formula (he squared the 0.1 uncertainty as though it were a part of the time) and he ended up with 20 with an uncertainty of .05 meters. Now this seems more realistic to me so im leaning toward him being right and me being wrong, the only thing i dont get is that if its possible for the object to travel in 2.1 s shouldnt the plus or minus have the 22.05 in its range? As that would be a viable answer? Another friend got 20.05 with .1 uncertainty but im not sure how he got his answer.

Any help would be appreciated. Thank you.

2. Sep 13, 2014

### Orodruin

Staff Emeritus
Your approach is essentially correct (although I would quote the best fit as 20 m). The error of the square is not the square of the error. In general
$$(A\pm dA)^2 = A^2 \pm 2 A\, dA + dA^2.$$
For small errors you can neglect the last term and the error is $2A\, dA$. If you had a 5% error in the original $A$, you will have ca 10% error in its square.

3. Sep 13, 2014

### psstudent

Ok thank you. So what im gathering is that I have the correct answer, but could you please elaborate why you would use 20 instead of 20.05 or are you just saying i should just round it down to 20 flat?

4. Sep 13, 2014

### Orodruin

Staff Emeritus
A combination of both. Your error is far larger than that precision and 20 m is what corresponds to the central value of the time measurement. Some times you will see asymmetric errors written as $20^{+2.05}_{-1.95}$ m, but again you do not really have enough significant digits (for example you have rounded g off to 10 m/s2). My answer would have been $20\pm 2$ m.