A students measures the distance traversed in | vedclass

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Question

From, the, relation
$h = ut = \frac{1}{2}g{t^2}$
$h = \frac{1}{2}g{t^2}$$ \Rightarrow g = \frac{{2h}}{{{t^2}}}$ (body, initially, at, rest)

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Vedclass · Accepted Answer

$e_1+2{e_2}$

Vedclass · Answer

From, the, relation
$h = ut = \frac{1}{2}g{t^2}$
$h = \frac{1}{2}g{t^2}$$ \Rightarrow g = \frac{{2h}}{{{t^2}}}$ (body, initially, at, rest)

Taking, natural, log aritham, on, both, sides, we, get

$In\,g = In\,h - 2\,In\,t$

Differentiating, $\frac{{\Delta h}}{g} = \frac{{\Delta h}}{h}\, - 2\,\frac{{\Delta t}}{t}$
For, max imum, permissible, error,

or,${\left( {\frac{{\Delta g}}{g} 	imes 100} ight)_{\max }} = \left( {\frac{{\Delta h}}{h} 	imes 100} ight) + 2 	imes \left( {\frac{{\Delta t}}{t} 	imes 100} ight)$

According, to, problem

$\frac{{\Delta h}}{h} 	imes 100{ = _{{e_1}}}\,and\,\frac{{\Delta t}}{t} 	imes 100{ = _{{e_2}}}$

Therefore, $( {\frac{{\Delta g}}{g} 	imes 100} )_{\max } = {e_1} + 2{e_2}$

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