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(B) For a body starting from rest,the distance $h$ covered in time $t$ is given by the equation of motion:$h = \frac{1}{2}gt^2$Rearranging for $g$,we

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$e_1 + 2e_2$

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(B) For a body starting from rest,the distance $h$ covered in time $t$ is given by the equation of motion:$h = \frac{1}{2}gt^2$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 to find the relative error:$\frac{\Delta g}{g} = \frac{\Delta h}{h} + 2\frac{\Delta t}{t}$For maximum permissible percentage error,we add the absolute values of the relative errors:$\left( \frac{\Delta g}{g} 	imes 100 ight)_{\max} = \left( \frac{\Delta h}{h} 	imes 100 ight) + 2 	imes \left( \frac{\Delta t}{t} 	imes 100 ight)$Given that $\frac{\Delta h}{h} 	imes 100 = e_1$ and $\frac{\Delta t}{t} 	imes 100 = e_2$,the percentage error in $g$ is:$	ext{Percentage error in } g = e_1 + 2e_2$

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