In the above problem, this horizontal distance between the two fragments when their position vectors are perpendicular to each other is

Options

A. 40 m
B. 20m
C. 10 m
D. 5m

Answer

To solve the problem of finding the horizontal distance between two fragments when their position vectors are perpendicular to each other, we can follow these steps:

### Step 1: Understand the Motion of the Fragments
The two fragments are projected horizontally with the same initial velocity. Let's denote the initial velocity as \( v = 10 \, \text{m/s} \). The fragments also experience vertical motion due to gravity.

### Step 2: Write the Position Vectors
For the two fragments, we can write their position vectors as:
- For the first fragment:
\[ \mathbf{r_1} = (10t) \hat{i} + \left(\frac{1}{2} g t^2\right) \hat{j} \]
- For the second fragment:
\[ \mathbf{r_2} = (10t) \hat{i} - \left(\frac{1}{2} g t^2\right) \hat{j} \]
where \( g \) is the acceleration due to gravity (approximately \( 9.8 \, \text{m/s}^2 \)).

### Step 3: Determine the Condition for Perpendicularity
The position vectors are perpendicular when their dot product is zero:
\[ \mathbf{r_1} \cdot \mathbf{r_2} = 0 \]
Calculating the dot product:
\[ (10t) \cdot (10t) + \left(\frac{1}{2} g t^2\right) \cdot \left(-\frac{1}{2} g t^2\right) = 0 \]
This simplifies to:
\[ 100t^2 - \frac{1}{4} g^2 t^4 = 0 \]

### Step 4: Factor the Equation
Factoring out \( t^2 \):
\[ t^2 \left(100 - \frac{1}{4} g^2 t^2\right) = 0 \]
This gives us two solutions: \( t^2 = 0 \) (which we discard) and:
\[ 100 - \frac{1}{4} g^2 t^2 = 0 \]

### Step 5: Solve for Time \( t \)
Rearranging gives:
\[ \frac{1}{4} g^2 t^2 = 100 \]
\[ g^2 t^2 = 400 \]
\[ t^2 = \frac{400}{g^2} \]
Taking the square root:
\[ t = \frac{20}{g} \]

### Step 6: Calculate the Horizontal Distance
The horizontal distance \( d \) traveled by the fragments when their position vectors are perpendicular is given by:
\[ d = v \cdot t = 10 \cdot t \]
Substituting \( t \):
\[ d = 10 \cdot \frac{20}{g} \]
Using \( g \approx 10 \, \text{m/s}^2 \) for simplicity:
\[ d = 10 \cdot 2 = 20 \, \text{m} \]

### Step 7: Final Calculation
Since both fragments are moving horizontally, the total horizontal distance between the two fragments is:
\[ \text{Total Distance} = 2 \cdot d = 40 \, \text{m} \]

### Conclusion
Thus, the horizontal distance between the two fragments when their position vectors are perpendicular to each other is \( 40 \, \text{meters} \).

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