Question

The distance travelled by a body falling freely from rest in $2^{nd},3^{rd}$ and $5^{th}$ second of its motion are in the ratio

• A. $1:3:5$
• B. $3:5:7$
• C. $3:5:9$
• D. $1:3:5$

Answer 1

The correct option is C $3:5:9$

As we know that the distance travelled in $n^{th}$ second is given as,
$S_n=u+\frac{a}{2}(2n-1)$ Where, $u=0,a=g=10m/s^2$
$S_n=0+\frac{a}{2}(2n-1)$
Distance travelled in $2^{nd}$ second
$S_2=\frac{g}{2}(2\times 2-1)$
Distance travelled in $3^{nd}$ second
$S_3=\frac{g}{2}(2\times 3-1)$
Distance travelled in $5^{nd}$ second
$S_5=\frac{g}{2}(2\times 5-1)$

$\therefore S_2:S_3:S_5=3:5:9$