Question

A stone falls freely from rest and the total distance
covered by it in the last second of its motion equals the
distance covered by it in the first three seconds of its
motion. The stone ramains in the air for:
(WB JEE 2010)
(a) 6 (b) 5s (c) 75
(d) 4s​

Answer 1

Answer:

Last sec Sn = a/2 (2n - 1) = 5(2n-1)

S in 3s= 1/2 .10.9 = 45

10n - 5 = 45

n = 5 s

Explanation:

Answer 2

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