Three stones A, B and C are simultaneously projected from same point with same speed. A is thrown upwards, B is thrown horizontally and C is thrown downwards from a building. When the distance between stone A and C becomes 10 m, then distance between A and B will be :

Options

A. 10m
B. 5m
C. `10sqrt(2) m
D. `5sqrt(2) m

Answer

To solve the problem, we need to analyze the motion of the three stones A, B, and C, which are thrown from the same point with the same initial speed but in different directions.

### Step-by-Step Solution:

1. **Identify the motion of each stone:**
- Stone A is thrown upwards.
- Stone B is thrown horizontally.
- Stone C is thrown downwards.

2. **Define the time of flight:**
Let the time of flight for all stones be \( t \).

3. **Calculate the vertical displacement of each stone:**
- For stone A (upwards):
\[ S_A = ut - \frac{1}{2}gt^2 \]
- For stone B (horizontal):
- Vertical displacement (due to gravity):
\[ S_{Bv} = \frac{1}{2}gt^2 \]
- Horizontal displacement:
\[ S_{Bh} = ut \]
- For stone C (downwards):
\[ S_C = ut + \frac{1}{2}gt^2 \]

4. **Determine the distance between stones A and C:**
Given that the distance between stones A and C is 10 m:
\[ |S_A - S_C| = 10 \]
Substituting the equations for \( S_A \) and \( S_C \):
\[ |(ut - \frac{1}{2}gt^2) - (ut + \frac{1}{2}gt^2)| = 10 \]
Simplifying this gives:
\[ | -gt^2 | = 10 \]
Thus:
\[ gt^2 = 10 \quad \text{(since distance cannot be negative)} \]

5. **Calculate the value of \( ut \):**
From the previous step, we have:
\[ ut = \frac{10}{g} \quad \text{(using } g = 10 \, \text{m/s}^2 \text{ for simplicity)} \]
Therefore:
\[ ut = 5 \, \text{m} \]

6. **Find the distance between stones A and B:**
The vertical distance of stone B from the original point is:
\[ S_{Bv} = \frac{1}{2}gt^2 = \frac{1}{2} \times 10 \times t^2 = 5 \, \text{m} \]
The horizontal distance of stone B is:
\[ S_{Bh} = ut = 5 \, \text{m} \]

7. **Calculate the resultant distance between A and B:**
Using the Pythagorean theorem:
\[ d_{AB} = \sqrt{(S_{Bh})^2 + (S_{Bv})^2} = \sqrt{(5)^2 + (5)^2} = \sqrt{25 + 25} = \sqrt{50} = 5\sqrt{2} \, \text{m} \]

### Final Answer:
The distance between stone A and stone B when the distance between stone A and stone C becomes 10 m is \( 5\sqrt{2} \, \text{m} \).