From the top of a tower, a particle is thrown vertically downwards with a velocity of $10 ext{ m/s}$. The ra

Question

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

Accepted Answer

1. **Identify the Formula:** The distance covered by a particle in the $n$-th second of uniformly accelerated motion is given by the formula:
   $$ S_n = u + \frac{a}{2}(2n - 1) $$
   where $u$ is initial velocity, $a$ is acceleration, and $n$ is the specific second .
2. **Identify Given Values:**
   - Initial velocity ($u$) = $10 	ext{ m/s}$ (downwards, taking down as positive).
   - Acceleration ($a$) = $g = 10 	ext{ m/s}^2$.
3. **Calculate Distance in 3rd Second ($S_3$):**
   Substitute $n=3$:
   $$ S_3 = 10 + \frac{10}{2}(2 	imes 3 - 1) $$
   $$ S_3 = 10 + 5(6 - 1) = 10 + 25 = 35 	ext{ m} $$
4. **Calculate Distance in 2nd Second ($S_2$):**
   Substitute $n=2$:
   $$ S_2 = 10 + \frac{10}{2}(2 	imes 2 - 1) $$
   $$ S_2 = 10 + 5(4 - 1) = 10 + 15 = 25 	ext{ m} $$
5. **Calculate Ratio:**
   $$ 	ext{Ratio} = \frac{S_3}{S_2} = \frac{35}{25} = \frac{7}{5} $$

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