[2010]
Solution:
For free fall, \(s = \frac{1}{2}gt^2 \Rightarrow g = \frac{2s}{t^2}\). The percentage error in (g) is given by the sum of percentage errors of (s) and twice the percentage error of (t). So, Percentage error in \(g = e_1 + 2e_2\).
A student measures the distance traversed in free fall of a body, initially at rest in a given time. He used this data to estimate (g), the acceleration due to gravity. If the maximum percentage errors in measurement of the distance and the time are \(e_1\) and \(e_2\) respectively, the percentage error in the estimation of (g) is:
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