In this experiment, the acceleration due to gravity is related tothe time taken

Question

In this experiment, the acceleration due to gravity is related tothe time taken t that it takes the ball to drop a height y throughthe formula g = 2y/t^2. Use the general approach for errorpropagation given next. In general if one has a quantity W= a^n *b^m * c^p, then the fractional error in W leading from the errorsa, b and c from the measurement of a, b, and c isgiven by:

W/W= sqrt((na/a)^2+(mb/b)^2+(pc/c)^2)

For the case g, show that the formula for the fractional errorg/g is:

g/g= sqrt((y/y)^2+(2t/t)^2)

Use the formulas for W/W, W and g given above


Please let me know if you need any moreinformation

Explanation / Answer

Using the formula that you have for W/W write yourfunction for g as: g=2*y1*t-2 This means that you now have it in the form W=an*bm with a=y b=t and n=1m=-2 Now using the formula that you have for W/W andsubstituting g/g g/g=(n*y/y)2+(m*t/t)2=(1*y/y)2+(-2*t/t)2=(y/y)2+(-1)2*(2*t/t)2 Which brings you to the final answer: g/g=(y/y)2+(2*t/t)2 because -12=1. All this comes about from using theformula for W/W Which brings you to the final answer: g/g=(y/y)2+(2*t/t)2 because -12=1. All this comes about from using theformula for W/W

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