A student measures the distance covered and time taken by a body...

From the relation

$\text{h} = \text{ut} + \dfrac{1}{2} {gt}^{2}$

\text{h} = \dfrac{1}{2} {gt}^{2} ⇒ \text{g} = \dfrac{2 \text{h}}{t^{2}} \left(∵ \text{body initially at rest}\right)<br/><br/>Taking natural logarithms on both sides, we get<br/><br/>\ln \left(g\right) = \ln \left(h\right) - 2\ln \left(t\right) + \ln \left(2\right)$<br/><br/>$\text{Differentiating,} $\dfrac{\Delta \text{g}}{\text{g}} = \dfrac{\Delta \text{h}}{\text{h}} -$2 \dfrac{\Delta \text{t}}{\text{t}}<br/><br/>For maximum permissible errors, we add both the errors. So<br/><br/> \left(\dfrac{\Delta \text{g}}{g} \times100\right)\text\max = \left(\dfrac{\Delta \text{h}}{h} \times \text{100}\right) +2 $\times$\left(\dfrac{\Delta \text{t}}{t} \times \text{100}\right)

\dfrac{\Delta \text{h}}{\text{h}} \times \text{100} = {e}_{1} \text{and} \dfrac{\Delta \text{t}}{\text{t}} \times \text{100} = {e}_{2}

\text{Therefore,} {\left(\dfrac{\Delta \text{g}}{\text{g}} \times \text{100}\right)}_\max = {e}_{1} +2 {e}_{2}