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(A) The distance $h$ covered by a body in free fall starting from rest in time $t$ is given by the equation $h = \frac{1}{2} gt^2$.Rearranging this fo

Vedclass · Accepted Answer

$e_1 + 2e_2$

Vedclass · Answer

(A) The distance $h$ covered by a body in free fall starting from rest in time $t$ is given by the equation $h = \frac{1}{2} gt^2$.Rearranging this formula to solve for $g$,we get $g = \frac{2h}{t^2}$.According to the rules of error propagation,for a quantity $g = k \cdot h^a \cdot t^b$,the relative error is given by $\frac{\Delta g}{g} = a \frac{\Delta h}{h} + |b| \frac{\Delta t}{t}$.Here,$a = 1$ and $b = -2$. Therefore,the maximum relative error is $\frac{\Delta g}{g} = \frac{\Delta h}{h} + 2 \frac{\Delta t}{t}$.Multiplying by $100$ to express this as a percentage error,we get $\frac{\Delta g}{g} 	imes 100 = \left( \frac{\Delta h}{h} 	imes 100 ight) + 2 \left( \frac{\Delta t}{t} 	imes 100 ight)$.Given that the percentage error in distance is $e_1$ and in time is $e_2$,the percentage error in $g$ is $e_1 + 2e_2$.

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