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(D) Given the formula for distance: $h = \frac{1}{2} gt^2$. Rearranging for $g$,we get: $g = \frac{2h}{t^2}$. Taking the natural logarithm on both sid

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$e_1+2e_2$

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(D) Given the formula for distance: $h = \frac{1}{2} gt^2$. Rearranging for $g$,we get: $g = \frac{2h}{t^2}$. Taking the natural logarithm on both sides: $\ln(g) = \ln(2) + \ln(h) - 2\ln(t)$. Differentiating both sides to find the relative error: $\frac{\Delta g}{g} = 0 + \frac{\Delta h}{h} + 2\frac{\Delta t}{t}$. For maximum percentage error,we add the absolute values of the relative errors: $\left(\frac{\Delta g}{g} 	imes 100ight)_{max} = \left(\frac{\Delta h}{h} 	imes 100ight) + 2\left(\frac{\Delta t}{t} 	imes 100ight)$. Given that the percentage error in $h$ is $e_1$ and in $t$ is $e_2$,we substitute these values: $	ext{Percentage error in } g = e_1 + 2e_2$.

If the maximum percentage error in the measurement of distance $(h)$ and time $(t)$ are '$e_1$' and '$e_2$' respectively,the percentage error in the measurement of acceleration of gravity $(g)$ is [use $h=\frac{1}{2} gt^2$].

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