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The ratio of the displacements of a freely falling body during first, second and third seconds of its motion is
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Verified Answer
The correct answer is:
$1: 3: 5$
Distance travelled in $\mathrm{n}^{\text {th }}$ second is given by
$$
\mathrm{S}_{\mathrm{n}}=\mathrm{u}+\frac{\mathrm{a}}{2}(2 \mathrm{n}-1)
$$
When a body is free falling $\mathrm{a}=\mathrm{g}$
Now,
Distance travelled in first second : $(n=1)$
$$
\mathrm{S}_1=0+\frac{\mathrm{g}}{2}(2-1)=\frac{\mathrm{g}}{2}
$$
Distance travelled in 2nd second: $(n=2)$
$$
\mathrm{S}_2=0+\frac{\mathrm{g}}{2}(2 \times 2-1)=\frac{3 \mathrm{~g}}{2}
$$
Distance travelled in 3rd second : $(n=3)$
$$
\mathrm{S}_3=0+\frac{\mathrm{g}}{2}(2 \times 3-1)=\frac{5 \mathrm{~g}}{2}
$$
Hence, the ratio is given by:
$$
\begin{aligned}
& \mathrm{S}_1: \mathrm{S}_2: \mathrm{S}_3=\frac{\mathrm{g}}{2}: \frac{3 \mathrm{~g}}{2}: \frac{5 \mathrm{~g}}{2} \\
& =1: 3: 5
\end{aligned}
$$
$$
\mathrm{S}_{\mathrm{n}}=\mathrm{u}+\frac{\mathrm{a}}{2}(2 \mathrm{n}-1)
$$
When a body is free falling $\mathrm{a}=\mathrm{g}$
Now,
Distance travelled in first second : $(n=1)$
$$
\mathrm{S}_1=0+\frac{\mathrm{g}}{2}(2-1)=\frac{\mathrm{g}}{2}
$$
Distance travelled in 2nd second: $(n=2)$
$$
\mathrm{S}_2=0+\frac{\mathrm{g}}{2}(2 \times 2-1)=\frac{3 \mathrm{~g}}{2}
$$
Distance travelled in 3rd second : $(n=3)$
$$
\mathrm{S}_3=0+\frac{\mathrm{g}}{2}(2 \times 3-1)=\frac{5 \mathrm{~g}}{2}
$$
Hence, the ratio is given by:
$$
\begin{aligned}
& \mathrm{S}_1: \mathrm{S}_2: \mathrm{S}_3=\frac{\mathrm{g}}{2}: \frac{3 \mathrm{~g}}{2}: \frac{5 \mathrm{~g}}{2} \\
& =1: 3: 5
\end{aligned}
$$
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