Question

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

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

Answer 1

The correct option is B $7:5$

$\frac{S_{3rd}}{S_{2nd}}=\frac{10+\frac{g}{2}(2\times 3-1)}{10+\frac{g}{2}(2\times 2-1)}[S_{nth}=u+\frac{a}{2}(2n-1\left)\right]\frac{S_{3rd}}{S_{2nd}}=\frac{10+5\left(5\right)}{10+5\left(3\right)}=\frac{35}{25}=\frac{7}{5}$