Question

The ratio of the distance travelled by a freely falling body in the 1^st, 2^nd, 3^rd and 4^th second

• A. 1:2:3:4
• B. 1:4:9:16
• C. 1:3:5:7
• D. 1:1:1:1

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
A "freely falling body" implies that the object is falling under the influence of gravity alone, starting from rest.
Thus, the initial velocity \(u = 0\) and the acceleration is constant (\(a = g\)).
The question asks for the distance travelled in specific time intervals (the \(n^{th}\) second).
Key Formula or Approach:
The distance travelled in the \(n^{th}\) second is given by the formula:
\[ S_n = u + \frac{a}{2}(2n - 1) \]
For a freely falling body starting from rest:
\[ S_n = 0 + \frac{g}{2}(2n - 1) \]
\[ S_n \propto (2n - 1) \]
Step 2: Detailed Explanation:
Let's calculate the distance for each required second:
- For the \(1^{st}\) second (\(n = 1\)):
\[ S_1 = \frac{g}{2}(2 \times 1 - 1) = \frac{g}{2}(1) \]
- For the \(2^{nd}\) second (\(n = 2\)):
\[ S_2 = \frac{g}{2}(2 \times 2 - 1) = \frac{g}{2}(3) \]
- For the \(3^{rd}\) second (\(n = 3\)):
\[ S_3 = \frac{g}{2}(2 \times 3 - 1) = \frac{g}{2}(5) \]
- For the \(4^{th}\) second (\(n = 4\)):
\[ S_4 = \frac{g}{2}(2 \times 4 - 1) = \frac{g}{2}(7) \]

The ratio is:
\[ S_1 : S_2 : S_3 : S_4 = \frac{g}{2}(1) : \frac{g}{2}(3) : \frac{g}{2}(5) : \frac{g}{2}(7) \]
\[ S_1 : S_2 : S_3 : S_4 = 1 : 3 : 5 : 7 \]
This is a famous result known as Galileo's law of odd numbers.
Step 3: Final Answer:
The ratio is \(1 : 3 : 5 : 7\).