Question

11. A student measures the distance traversed in free
fall of a body, initially at rest in a given time. He
uses this data to estimate g, the acceleration due
to gravity. If the maximum percentage errors in
measurement of the distance and the time are e,
and e, respectively, the percentage error in the
estimation of g is
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Answer 1

here is ur answer of ur questions

Answer 2

From the formula we know that, $\mathrm{h}=\mathrm{ut}+1 / 2 \mathrm{gt}^{2}$, body is initially at rest so ‘’u’’ will be zero.  

$\begin{array}{l}{h=1 / 2 g t^{2}} \\ {g=2 h / t^{2}}\end{array}$

Now take logarithm on both sides

log g = log h – log t after differentiating this equation we will get  

Δg/g = Δh/h – 2. Δt /t

For maximum error, (Δg/g x 100) max = (Δh/h x 100) + 2 (Δt/t x 100)

According to given question equation will be as follows,

(Δh/h x 100) = e1, (Δt/t x 100) = e and (Δg/g x 100) max = $\mathrm{e}_{1}+2 \mathrm{e}_{2}$