A body is thrown up with a velocity `100ms^(-1)` it travels 5 m in the last swecond of upward journey if the same body thrown up with velocity `200ms^(-1)`, how much distance (in metre) will it travel in the last second of its upward journey `(g=10ms^(-2))`

Explanation

To solve the problem, we need to determine the distance traveled by a body thrown upwards with an initial velocity of \(200 \, \text{m/s}\) in the last second of its upward journey, given that the acceleration due to gravity \(g = 10 \, \text{m/s}^2\).

### Step-by-Step Solution:

1. **Understanding the Motion**:
When a body is thrown upwards, it will eventually stop and then start falling back down due to gravity. The distance traveled in the last second of the upward journey can be calculated using the kinematic equations of motion.

2. **Using the Given Information**:
From the problem, we know that when the body is thrown with an initial velocity of \(100 \, \text{m/s}\), it travels \(5 \, \text{m}\) in the last second of its upward journey.

3. **Finding the Time of Flight**:
The time taken to reach the maximum height can be calculated using the formula:
\[ t = \frac{u}{g} \]
where \(u\) is the initial velocity. For \(u = 100 \, \text{m/s}\):
\[ t = \frac{100}{10} = 10 \, \text{s} \]
Thus, the total time of flight for the first case is \(10 \, \text{s}\).

4. **Distance in the Last Second**:
The distance traveled in the last second can be calculated using the formula:
\[ s = u + \frac{1}{2} a t^2 \]
However, since we know the distance traveled in the last second is \(5 \, \text{m}\), we can use the relationship for the last second:
\[ s = v_1 - v_0 \]
where \(v_1\) is the final velocity at the end of the last second and \(v_0\) is the velocity at the beginning of the last second.

5. **Calculating the Final Velocity**:
The final velocity at the top of the motion (when the body stops) is \(0\). The initial velocity at the start of the last second (which is \(t-1\)) can be found using:
\[ v = u - gt \]
For \(t = 9 \, \text{s}\):
\[ v_0 = 100 - 10 \times 9 = 10 \, \text{m/s} \]
Thus, the distance traveled in the last second is:
\[ s = 10 - 0 = 10 \, \text{m} \]

6. **Applying the Same Logic for \(200 \, \text{m/s}\)**:
Now, we repeat the process for the initial velocity of \(200 \, \text{m/s}\):
\[ t = \frac{200}{10} = 20 \, \text{s} \]
The initial velocity at the start of the last second (which is \(t-1 = 19 \, \text{s}\)):
\[ v_0 = 200 - 10 \times 19 = 10 \, \text{m/s} \]
The distance traveled in the last second is:
\[ s = 10 - 0 = 10 \, \text{m} \]

### Final Answer:
The distance traveled in the last second of the upward journey when thrown with an initial velocity of \(200 \, \text{m/s}\) is **10 meters**.