If a particle is thrown vertically upwards , then its velocity so that it covers same distance in ` 5th and 6th seconds would be .

Options

A. (a) ` 48 m//s`
B. (b) `14 m//s`
C. (c ) ` 49 m//s`
D. (d) ` 7 m//s`

Explanation

To solve the problem of finding the initial velocity of a particle thrown vertically upwards so that it covers the same distance in the 5th and 6th seconds, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Motion**:
- When a particle is thrown upwards, it will rise until it reaches its highest point, where its velocity becomes zero, and then it will start descending. The distance covered in the 5th and 6th seconds being the same implies that the particle reaches its highest point at the end of the 5th second.

2. **Time to Reach the Highest Point**:
- Since the particle reaches its highest point at the end of the 5th second, it means that the time taken to reach the highest point is 5 seconds.

3. **Using the Equation of Motion**:
- We can use the first equation of motion:
\[ V = U + at \]
where:
- \( V \) is the final velocity (0 m/s at the highest point),
- \( U \) is the initial velocity,
- \( a \) is the acceleration (which is \(-g\), where \( g \approx 9.8 \, \text{m/s}^2 \)),
- \( t \) is the time (5 seconds in this case).

4. **Setting Up the Equation**:
- Plugging in the values, we have:
\[ 0 = U - g \cdot 5 \]
- Rearranging gives:
\[ U = g \cdot 5 \]

5. **Calculating the Initial Velocity**:
- Substituting \( g = 9.8 \, \text{m/s}^2 \):
\[ U = 9.8 \cdot 5 = 49 \, \text{m/s} \]

6. **Conclusion**:
- Therefore, the initial velocity required for the particle to cover the same distance in the 5th and 6th seconds is \( 49 \, \text{m/s} \).

### Final Answer:
The initial velocity \( U \) required is \( 49 \, \text{m/s} \).