Question

The distance moved by a freely falling body during the $1^{st},2^{nd}$ and $3^{rd}$ seconds of its motion are proportional to :

• A. 1 : 2 : 3
• B. 1 : 3 : 5
• C. 1 : 4 : 9
• D. 1 : 1 : 1

Answer 1

The correct option is B 1 : 3 : 5

Initial velocity of the body $u=0$

Distance covered in $n^{th}$ second, $S_n=u+\frac{1}{2}g(2n-1)$

$\therefore$ $S_n=\frac{1}{2}g(2n-1)$ (as $u=0$)

Distance moved in $1^{st}$ second i.e. $n=1$, $S_1=\frac{1}{2}g(2\times 1-1)=\frac{1}{2}g$

Distance moved in $2^{nd}$ second i.e. $n=2$, $S_2=\frac{1}{2}g(2\times 2-1)=\frac{3}{2}g$

Distance moved in $3^{rd}$ second i.e. $n=3$, $S_3=\frac{1}{2}g(2\times 3-1)=\frac{5}{2}g$

$⟹$ $S_1:S_2:S_3=1:3:5$