A body is thrown up with a velocity `1000 m s^(-1)`. It travels `5 m` in the last second of its journey. If the same body is thrown up with a velocity `200 m s^(-1)`. How much distance (in metre) will it travel in the last second (g= 10 m s^(-2))?.

Explanation

To solve the problem, we need to analyze the motion of the body thrown upwards and how much distance it travels in the last second of its journey when thrown with different initial velocities.

### Step-by-Step Solution:

1. **Understanding the Motion:**
When a body is thrown upwards, it decelerates due to gravity until it reaches its maximum height, after which it falls back down. The distance traveled in the last second of its upward journey is equal to the distance traveled in the first second of its downward journey due to symmetry.

2. **Given Data:**
- Initial velocity \( u = 1000 \, \text{m/s} \)
- Distance traveled in the last second of the upward journey \( s = 5 \, \text{m} \)
- Acceleration due to gravity \( g = 10 \, \text{m/s}^2 \)

3. **Finding the Time of Flight for the First Case:**
The distance traveled during the last second of the upward journey can be expressed as:
\[ s = u - \frac{g}{2} \cdot (t - 1) \]
where \( t \) is the total time of flight. Rearranging gives:
\[ 5 = 1000 - \frac{10}{2} \cdot (t - 1) \]
Simplifying this:
\[ 5 = 1000 - 5(t - 1) \]
\[ 5 = 1000 - 5t + 5 \]
\[ 0 = 1000 - 5t \]
\[ 5t = 1000 \]
\[ t = 200 \, \text{s} \]

4. **Using the Time of Flight for the Second Case:**
Now, we need to find the distance traveled in the last second of the upward journey when the initial velocity is \( u = 200 \, \text{m/s} \).

5. **Finding the Time of Flight for the Second Case:**
Using the same formula for the last second of the upward journey:
\[ s = u - \frac{g}{2} \cdot (t - 1) \]
We need to find \( t \) for \( u = 200 \, \text{m/s} \):
\[ s = 200 - \frac{10}{2} \cdot (t - 1) \]
We can express the distance traveled in the last second as:
\[ s = 200 - 5(t - 1) \]
We know that the total time \( t \) can be calculated from the formula for the time of flight:
\[ t = \frac{2u}{g} = \frac{2 \times 200}{10} = 40 \, \text{s} \]

6. **Calculating the Distance in the Last Second:**
Now we can plug this \( t \) back into the distance formula:
\[ s = 200 - 5(40 - 1) \]
\[ s = 200 - 5(39) \]
\[ s = 200 - 195 = 5 \, \text{m} \]

### Final Answer:
The distance traveled in the last second when the body is thrown with a velocity of \( 200 \, \text{m/s} \) is **5 meters**.