Question

what is the distance covered by the freely falling body during the first 3 seconds of the motion

Answer 1

$\\ g = 10m/{s}^{-2} \\ \\ t = 3s \\ \\ u = 0m/{s}^{-1} \\ \\ s = ut + \frac {1}{2} g{t}^{2} \\ \\ s = 0 \times 3 + \frac {1}{2} \times 10 \times 9 \\ \\ s = 5 \times 9 \\ \\ Distance \: Travelled = 45 m$

Answer 2

$\fcolorbox{red}{white}{solution : - } \\ \\ \underline{ \red{we \: know \: the \: acceleration \: due \: to \:}} \\ \underline{ \red{gravity \: is \: }9.8 \: m/ {s}^{2} \:\ \red{ but \: we \: take \: }10 \: m/ {s}^{2} } \\ \\\underline{ \red {now \: }} \\ \underline{to \: calculate \: distance \: travel \: by \: body} \\ \underline{ we \: use}\\ \mathfrak{ \red{formula : - }} \\ \\ s = ut + \frac{1}{2} a {t}^{2} \\ \\ \underline \red{where} \\ \\ {\red{v \:} = \: final \: velocity \: } \\ {\red{u \: } = \: initial \: velocity \: } \\ {\red{a \:} = \: acceleration} \\ {\red{t \:} = \: time \: taken} \\ \\ \underline \red{we \: know \: initial \: velocity \: is \: 0} \ \\ \underline{ hence} \\ \\ \\ s = \frac{1}{2} \times 10 \times {3}^{2} \\ \\ s \: = \frac{1}{2} \times 10 \times 9 \\ \\ s = \frac{90}{2} = 45 \: meter \\ \\ \underline{\fbox{to \: find}} \\ \\ \underline{ \: How \: long \: does \: the \: body \: remain \: in \: air ... } \\ \\\underline{ \red{ we \: use}} \\ \\ s = \frac{u + a}{2} \times (2n - 1) \\ \\ we \: know \: \red { u = 0 } \: and \: \: \red{ a = 10} \: and \: \red{ s = 45 \: } \: \\ so \\ \\ 45 = \frac{0 + 10}{2} \times ( 2n - 1) \\ \\ \\ 45 = 5(2n - 1) \\ i.e. \\ \\ 5(2n - 1) = 45 \\ \\ 2n - 1 = \frac{45}{5} \\ \\ 2n - 1 = 9 \\ \\ 2n = 9 + 1 \\ \\ 2n = 10 \\ \\ n \: = \frac{10}{2} \\ \\ \fcolorbox{red}{white}{ n \: = 5 \: sec} \\ \\ \underline{ so \: we \: know \: now \: } \\ \\ \small{\fcolorbox{red}{white}{The \: body \: remains \: in \: the \: air \: for \: 5 \: sec}} \\ \\ \underline\red{i \: hope \: it \: helps \: you}.....$