Question

From the top of a tower, a particle is thrown vertically downwards with a velocity of 10 m/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

Answer: (B)

7 : 5

Answer 2

$S_{3^{rd}} = 10 + \frac{10}{2}(2 \times 3 - 1) = 35\mspace{6mu} m$

$S_{2^{nd}} = 10 + \frac{10}{2}(2 \times 2 - 1) = 25m$ ⇒ $\frac{S_{3^{rd}}}{S_{2^{nd}}} = \frac{7}{5}$