If the distance travelled by a freely falling body in the last second of its journey is equal to the distance travelled in the first 2s, the time of descent of the body is

Options

A. 5 s
B. 1.5 s
C. 2.5 s
D. 3 s

Answer

To solve the problem, we need to find the time of descent \( t_0 \) of a freely falling body, given that the distance traveled in the last second of its journey is equal to the distance traveled in the first 2 seconds.

### Step-by-Step Solution:

1. **Understanding the Motion**:
- A freely falling body is under constant acceleration due to gravity \( g \).
- The body is dropped from rest, so its initial velocity \( u = 0 \).

2. **Distance Traveled in the Last Second**:
- The distance traveled in the last second of the journey can be calculated using the formula for distance traveled in the \( n \)-th second:
\[ s_n = u + \frac{1}{2} g (2n - 1) \]
- Since \( u = 0 \) and \( n = t_0 \) (the total time of descent), the formula simplifies to:
\[ s_{last} = \frac{1}{2} g (2t_0 - 1) \]

3. **Distance Traveled in the First 2 Seconds**:
- The distance traveled in the first 2 seconds is given by:
\[ s_{first\ 2\ seconds} = \frac{1}{2} g t^2 \]
- For \( t = 2 \):
\[ s_{first\ 2\ seconds} = \frac{1}{2} g (2^2) = \frac{1}{2} g \cdot 4 = 2g \]

4. **Setting Up the Equation**:
- According to the problem, the distance traveled in the last second is equal to the distance traveled in the first 2 seconds:
\[ \frac{1}{2} g (2t_0 - 1) = 2g \]

5. **Solving the Equation**:
- Cancel \( g \) from both sides (assuming \( g \neq 0 \)):
\[ \frac{1}{2} (2t_0 - 1) = 2 \]
- Multiply both sides by 2:
\[ 2t_0 - 1 = 4 \]
- Add 1 to both sides:
\[ 2t_0 = 5 \]
- Divide by 2:
\[ t_0 = \frac{5}{2} \text{ seconds} \]

### Final Answer:
The time of descent of the body is \( t_0 = \frac{5}{2} \) seconds.

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