# Help with error analysis! ly needed

1. Oct 27, 2013

### slasakai

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. Oct 27, 2013

### 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.

3. Oct 27, 2013

### haruspex

That looks like the error (in degrees) of theta. That's not the error in sin theta. Need to convert to radians.

4. Oct 28, 2013

### slasakai

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