A simple uncertainty question about height.

[psstudent]

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

The questions reads like this:
"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 I am 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)mNow 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 that's 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 I am leaning toward him being right and me being wrong, the only thing i don't get is that if its possible for the object to travel in 2.1 s shouldn't 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 I am not sure how he got his answer.Any help would be appreciated. Thank you.

 

[Orodruin]

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.

 

[psstudent]

Ok thank you. So what I am 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?

 

[Orodruin]

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.

 

[HalfLostAndSearching]

I would suggest approaching this question by first identifying the key variables and their uncertainties. In this case, the variables are time (t) and acceleration due to gravity (a). The given time has an uncertainty of ±0.1 s, while the acceleration due to gravity is a known constant of 10 m/s^2.

Next, I would recommend using error propagation methods to calculate the uncertainty in the final answer. This involves propagating the uncertainties from the initial variables to the final answer using mathematical operations. In this case, since we are using the formula S= ut + 0.5at^2, we can use the following formula for error propagation: ΔS = |uΔt + 0.5aΔt^2|. This will give us the uncertainty in the final answer, which we can then add/subtract from the calculated value to get the range of possible values.

Using this method, I calculated the height to be 20.05 ± 0.05 m, which is consistent with your friend's answer. This means that the actual height could be anywhere between 20.00 m and 20.10 m.

Regarding your concern about the large uncertainty of 2 m, it is important to note that the uncertainty in the time (±0.1 s) is the main contributor to this large range. This is because we are squaring the time in the formula, which amplifies the uncertainty. However, this is a realistic uncertainty given the precision of the measurement of time.

In conclusion, it is important to carefully consider the uncertainties in all variables and use error propagation methods to calculate the uncertainty in the final answer. This will give a more accurate and realistic range of values for the final answer. I hope this helps clarify the solution to the question.