A body is thrown vertically upwards and takes 5 seconds to reach maximum height. The distance travelled by the body will be same in :-

Options

A. `1^(st) and 10^(th)` second
B. `2^(st) and 8^(th)` second
C. `4^(st) and 6^(th)` second
D. Both (2) & (3)

Answer

To solve the problem step by step, we need to analyze the motion of the body thrown vertically upwards and determine the distance traveled at different time intervals.

### Step-by-Step Solution:

1. **Understanding the Motion**:
- A body is thrown vertically upwards and takes 5 seconds to reach its maximum height.
- At maximum height, the velocity of the body is 0 m/s.

2. **Total Time of Flight**:
- Since the time taken to reach maximum height is 5 seconds, the total time of flight (up and down) will be:
\[ T = 5 \, \text{s (up)} + 5 \, \text{s (down)} = 10 \, \text{s} \]

3. **Distance Traveled in the First 1 Second**:
- Let the initial velocity be \( u \) and the acceleration due to gravity be \( g \) (approximately \( 9.8 \, \text{m/s}^2 \)).
- The distance traveled in the first second (from \( t = 0 \) to \( t = 1 \)) can be calculated using the equation of motion:
\[ s_1 = ut - \frac{1}{2} g t^2 \]
- For \( t = 1 \):
\[ s_1 = u(1) - \frac{1}{2} g (1^2) = u - \frac{g}{2} \]

4. **Distance Traveled in the Last 1 Second (from 9 to 10 seconds)**:
- The distance traveled in the last second (from \( t = 9 \) to \( t = 10 \)) can be calculated similarly:
\[ s_{10} = u(10) - \frac{1}{2} g (10^2) - \left( u(9) - \frac{1}{2} g (9^2) \right) \]
- We can simplify this using the symmetry of the motion:
\[ s_{10} = u(10) - \frac{1}{2} g (100) - \left( u(9) - \frac{1}{2} g (81) \right) \]
- Since the body returns to the same height, we can conclude:
\[ s_{10} = u - \frac{g}{2} \]

5. **Comparing Distances**:
- From the calculations, we see that:
\[ s_1 = s_{10} \]
- Thus, the distance traveled in the first second is equal to the distance traveled in the last second.

6. **Conclusion**:
- The distance traveled by the body will be the same in the first second and the last second (10 seconds).