Question

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

• A. 5:07
• B. 7:05
• C. 3:06
• D. 6:03

Answer 1

Answer: (B)

7:05

Answer 2

$s_{3rd}=10+\frac{10}{2}(2 \times 3 -1 )=35 m$ $s_{2nd}=10+\frac{10}{2}(2 \times 2-1 )=25 m$ $\Rightarrow$ $\frac{s_{2nd}}{s_{3rd}}=\frac{7}{5}$