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 remains in the
air for
(A) 3 sec
(B) 5 sec
(C) 7 sec
(D) 4 sec​

Answer 1

$\huge\blue{Hello\:FRIEND}$

HERE IS YOUR ANSWER

Answer is = 5 sec

Option (A) is right......

SEE THE ATTACHED PIC

MAY THIS HELPS YOU ✌️

$\huge\red{ཋཛ\:ཋrainly}$

Answer 2

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