Explanation:
B Since, stone falls freely, hence its initial velocity, \(u=0\)
Distance covered by the stone in first \(5 \mathrm{~s}\), is given as
\(\mathrm{h}_{1} =\mathrm{ut}+\frac{1}{2} \mathrm{gt}^{2}\)
\(=0 \times 5+\frac{1}{2} \times 9.8 \times 5^{2}\)
\(=122.5\)
Since, distance travelled in last second \(\left(\mathrm{t}^{\text {th }}\right.\) second) is equal to distance travelled in first \(5 \mathrm{~s}\), hence
\(\mathrm{h}_{1}=\mathrm{u}+\frac{1}{2} \mathrm{~g}(2 \mathrm{t}-1)\)
\(122.5=0+\frac{1}{2} \times 9.8(2 \mathrm{t}-1)\)
\(2 \mathrm{t}-1=\frac{122.5 \times 2}{9.8}\)
\(2 \mathrm{t}-1=25 \Rightarrow 2 \mathrm{t}=25+1\)
\(2 \mathrm{t}=26 \Rightarrow \mathrm{t}=13 \mathrm{sec}\)