A stone is dropped from a certain height which can reach the ground in $5 \mathrm{~s}$. It is stopped after $3 \mathrm{~s}$ of its fall and then it again released. The total time taken by the stone to reach the ground will be

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Aicharya · Accepted Answer

Total height to fall
$h=\frac{1}{2} \mathrm{gt}^2=\frac{1}{2} \times 10 \times(5)^2=125 m$
In first 3 sec.
$s_1=\frac{1}{2} \mathrm{gt}_1^2=\frac{1}{2} \times 10(3)^2=45 \mathrm{~m}$
Now remaining height to fall
$\therefore s_2=h-s_1=125-45=80 m$
$t_2=\sqrt{\frac{2 s_2}{g}}=\sqrt{\frac{2 \times 80}{10}}=4 \mathrm{sec}$
$t=t_1+t_2=3+4=7 \mathrm{sec}$

A stone is dropped from a certain height which can reach the ground in \(5 \mathrm{~s}\). It is stopped…

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