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A student measures the distance traversed in free fall of a body, initially at rest in a given time. He uses 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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The correct answer is:
$e_1+2 e_2$
$$
\begin{aligned}
& h=\frac{1}{2} g t^2 \\
& \text { or } \quad g=\frac{\mathrm{h}}{\mathrm{t}^2} \\
& \therefore \quad \log \mathrm{g}=\log \mathrm{h}-2 \log \mathrm{t} \\
& \left(\frac{\Delta g}{g} \times 100\right)_{\max }=\left(\frac{\Delta \mathrm{h}}{\mathrm{h}} \times 100\right)+2\left(\frac{\Delta \mathrm{t}}{\mathrm{t}} \times 100\right) \\
& =\mathrm{e}_1+2 \mathrm{e}_2 \\
&
\end{aligned}
$$
\begin{aligned}
& h=\frac{1}{2} g t^2 \\
& \text { or } \quad g=\frac{\mathrm{h}}{\mathrm{t}^2} \\
& \therefore \quad \log \mathrm{g}=\log \mathrm{h}-2 \log \mathrm{t} \\
& \left(\frac{\Delta g}{g} \times 100\right)_{\max }=\left(\frac{\Delta \mathrm{h}}{\mathrm{h}} \times 100\right)+2\left(\frac{\Delta \mathrm{t}}{\mathrm{t}} \times 100\right) \\
& =\mathrm{e}_1+2 \mathrm{e}_2 \\
&
\end{aligned}
$$
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