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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
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Verified Answer
The correct answer is:
$5 \mathrm{~s}$
Hints : $u=0$
$$
\begin{aligned}
& \mathrm{S}_3=0+\frac{1}{2} g t^2=\frac{1}{2} \times 10 \times 9=45 \\
& \mathrm{~S}_t \text { th }=u+(2 t-1) \frac{g}{2} \\
& \mathrm{~S}_t \text { th }=0+5(2 t-1)=45 \\
& 2 t-1=9 \\
& t=5 \mathrm{sec}
\end{aligned}
$$
$$
\begin{aligned}
& \mathrm{S}_3=0+\frac{1}{2} g t^2=\frac{1}{2} \times 10 \times 9=45 \\
& \mathrm{~S}_t \text { th }=u+(2 t-1) \frac{g}{2} \\
& \mathrm{~S}_t \text { th }=0+5(2 t-1)=45 \\
& 2 t-1=9 \\
& t=5 \mathrm{sec}
\end{aligned}
$$
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