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Q.
The distances covered by a freely falling body in its first, second, third,..... $n^{th}$ seconds of its motion
Motion in a Straight Line
Solution:
Distance travelled by a body in $n^{th}$ second is $D_{n}=u+\frac{a}{2}\left(2n-1\right)$
Here, $u = 0$, $a=g$
$\therefore $ Distance travelled by the body in $1^{st}$ second is $D_{1}=0+\frac{g}{2}\left(2\times1-1\right)=\frac{1}{2}\,g$
Distance travelled by the body in $2^{nd}$ second is $D_{2}=0+\frac{g}{2}\left(2\times2-1\right)=\frac{3}{2}g$
Distance travelled by the body in $3^{rd}$ second is
$D_{3}=0+\frac{g}{2}\left(2\times3-1\right)=\frac{5}{2}g$
and so on.
Hence, the distances covered by a freely falling body in its first, second, third.....$n^{th}$ second of its motion forms an arithmetic progression.