Question

The distance through which a body falls in the $n^{th}$ second is $h$. The distance through which it falls in the next second is

• A. $h$
• B. $h+\frac{g}{2}$
• C. $h-g$
• D. $h+g$

Answer 1

The correct option is D $h+g$

Wkt, according to $2^{rd}$ Kinematic equation ,
$S=ut+\frac{1}{2}a{t}^2$
Distance in $n^{th}sec=\frac{1}{2}g{n}^2$
Distance in ${(n+1)}^{th}sec=\frac{1}{2}g{(n+1)}^2$
Distance through which it falls in $n^{th}sec$ be $y$
$y=$ Distance in${(n+1)}^{th}sec-$ Distance in $n^{th}sec\Rightarrow \frac{1}{2}g{(n+1)}^2-\frac{1}{2}g{n}^2\Rightarrow \frac{g}{2}(2n+1)-----\left(1\right)$
Also, WKT distance in $n^{th}sec=S_n=u+\frac{a}{2}(2n-1)\Rightarrow S_n=\frac{g}{2}(2n-1)----\left(2\right)$
From $\left(1\right)$ and $\left(2\right)$
$y=h+g$