A body is thrown up with a velocity `100ms^(-1)`. It travels 5 m in the last second 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))`

Options

A. 5m
B. 10 m
C. 15 m
D. 20 m

Answer

To solve the problem step by step, we need to analyze the motion of the body thrown upwards and apply the equations of motion.

### Step 1: Understand the given data
We are given:
- Initial velocity \( u_1 = 100 \, \text{m/s} \)
- Distance traveled in the last second of the upward journey \( s_1 = 5 \, \text{m} \)
- Acceleration due to gravity \( g = 10 \, \text{m/s}^2 \)

### Step 2: Calculate the time taken to reach the maximum height for the first case
Using the formula for distance traveled in the last second of motion:
\[ s = u + \frac{1}{2} a t \]
For the last second of the journey, we can rearrange the formula to find the time \( t \) it takes to reach the maximum height.

The distance traveled in the last second can also be expressed as:
\[ s_1 = u_1 - \frac{1}{2} g (t - 1) - (u_1 - g t) \]
This simplifies to:
\[ s_1 = (u_1 - g t) + \frac{1}{2} g \]

### Step 3: Find the time \( t_1 \) for the first case
From the first case, we know:
\[ 5 = (100 - 10t_1) + 5 \]
This simplifies to:
\[ 0 = 100 - 10t_1 \]
Thus, solving for \( t_1 \):
\[ 10t_1 = 100 \implies t_1 = 10 \, \text{s} \]

### Step 4: Calculate the distance traveled in the last second for the second case
Now, we need to analyze the second scenario where the body is thrown with an initial velocity \( u_2 = 200 \, \text{m/s} \).

### Step 5: Find the time \( t_2 \) for the second case
Using the same approach, we can find the time \( t_2 \) for the second case:
\[ s_2 = u_2 - g(t_2 - 1) - (u_2 - g t_2) \]
This gives us:
\[ s_2 = (200 - 10t_2) + 5 \]

### Step 6: Solve for \( t_2 \)
Using the same logic as before:
\[ s_2 = (200 - 10t_2) + 5 \]
This simplifies to:
\[ s_2 = 205 - 10t_2 \]
We need to find \( t_2 \) when the body reaches the maximum height. The time taken to reach the maximum height can be calculated as:
\[ t_2 = \frac{u_2}{g} = \frac{200}{10} = 20 \, \text{s} \]

### Step 7: Calculate the distance traveled in the last second of the second case
Now substituting \( t_2 = 20 \, \text{s} \) into the distance formula:
\[ s_2 = (200 - 10 \cdot 20) + 5 = (200 - 200) + 5 = 5 \, \text{m} \]

### Final Answer
Thus, the distance traveled in the last second of the upward journey when thrown with a velocity of \( 200 \, \text{m/s} \) is:
\[ \boxed{5 \, \text{m}} \]