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Help with error analysis! ly needed

  1. Oct 27, 2013 #1
    Help with error analysis! urgently needed

    1. The problem statement, all variables and given/known data
    Using the theoretical prediction of t, find the errors of both t and x, . The problem is based on an experiment I did in Labs, In which a small metal ball was fired at 40° (uncertainty 0.5°) to the horizontal, at a height,y= 1.095m (with uncertainty of 0.005m) and fired at another measured quantity of initial speed V=5.23 m/s(uncertainty of 0.01m/s). and t is the time taken to hit the ground and x is the range of the ball.

    2. Relevant equations

    y=y0 - (Vsinθ)t - (0.5g)t^2

    using quadratic equation to solve for t

    t=((Vsinθ+ sqrt((Vsinθ)^2 + (2gy))/g

    and later plug in value of t
    into equation x=(Vcos40)t for x

    also we are instructed to ignore the uncertainty on the constant g
    3. The attempt at a solution

    putting in values t=0.926 s(3sf) and although I cannot type all of my working here, I used a series of steps calculating the errors on individual functions of the equation for t and slowly combining them, however it keeps resulting in an error of 1.17616421..... which obviously cannot be correct as this is larger than the predicted value! I'm clearly making a mistake somewhere, a solution will be greatly appreciated.

    Thanks gneill, I see how not including any working would make it difficult to find my mistake. I will try my best to describe the method I used here.

    I used the functional approach to calculate all the errors, as I couldn't really understand the calculus method.

    here goes:

    1. I first calculated the error on sinθ, using α=abs(cosθ) * error on sinθ = 0.3830222216

    2. Then the error on Vsinθ, using α=sqrt((errorV/V)^2 + (errorsinθ/sinθ)^2) * Vsinθ = 2.003216532

    3. Then the error on (Vsinθ)^2 , α=abs(2*Vsinθ) * (error on Vsinθ) = 13.4687433 [this value is the one where I start to doubt myself]

    4. Then I calculate the error on 2gy using α= abs(2g) * (error on value of y) = 0.0981
    [wasnt completely sure about taking g to be a constant despite being told to ignore the uncertainty]

    5. combining errors of the (Vsinθ)^2 + (2gy) using α= sqrt( errora^2 + errorb^2) = 13.46910059 [an obviously massive value]

    6. propogating this error through a power of 0.5, using same functional approach = 1.176164221

    7. finally adding the errors of everything under the sqrt sign and the Vsinθ outside it using

    α= sqrt( errora^2 + errorb^2) gives such a ridiculous answer = appx 2.3, which is so wrong....

    I know this was a bad way to write it out, but if anyone can follow it I would greatly appreciate it
    Last edited: Oct 27, 2013
  2. jcsd
  3. Oct 27, 2013 #2


    User Avatar

    Staff: Mentor

    Hi slasakai. Welcome to Physics Forums.

    There's not much we can do to find your error without seeing the work. But I can tell you that the result for the error in t should be closer to 0.001 seconds than 1 s.

    If you're proficient with calculus and taking derivatives, you can find the error for the whole expression at once via:

    $$\delta t = \sqrt{\left(\frac{df}{d \theta}\right)^2\delta \theta^2 +
    \left(\frac{df}{dv}\right)^2\delta v^2 +
    \left(\frac{df}{dy}\right)^2\delta y^2 }$$
    where ##f(\theta, v, y)## is the function for the time that you found and the derivatives are taken to be partial derivatives. The ##\delta \theta, \delta v, ## and ##\delta y## are the measurement errors.
  4. Oct 27, 2013 #3


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    That looks like the error (in degrees) of theta. That's not the error in sin theta. Need to convert to radians.
  5. Oct 28, 2013 #4
    thank you so mcuh haruspex, you have literally helped me so much by spotting that error. It was driving me nuts trying to figure out where I'd one wrong
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