Question

From the top of a tower, a particle is thrown vertically downwards with a velocity of $10m/s$. The ratio of distances covered by it in the $3^{rd}$ and $2^{nd}$ seconds of its motion is (Take $g=10m/s^2$)

• A. $5:7$
• B. $7:5$
• C. $3:5$
• D. $6:3$

Answer 1

The correct option is B $7:5$

We know that distance covered by the particle at any $n^{th}$ second is given by
$S_n=u+\frac{1}{2}a(2n-1)$
$\Rightarrow S_{3^{rd}}=10+\frac{10}{2}(2\times 3-1)=35m$
$\Rightarrow S_{2^{nd}}=10+\frac{10}{2}(2\times 2-1)=25m$
$\therefore \frac{S_{3^{rd}}}{S_{2^{nd}}}=\frac{7}{5}$