In the above problem, the time taken by the displacement vectors of the two fragments to become perpendicular to each other is

Options

A. 1s
B. 1.5s
C. 2s
D. 4s

Answer

To solve the problem of finding the time taken by the displacement vectors of two fragments to become perpendicular to each other, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We need to find the time at which the displacement vectors of two fragments become perpendicular. This means the angle between the two displacement vectors will be 90 degrees.

2. **Setting Up the Scenario**:
- Assume two fragments are projected from the same point with certain initial velocities. Let’s denote the horizontal component of the velocity as \( v_x \) and the vertical component as \( v_y \).

3. **Using the Kinematic Equations**:
- The horizontal displacement \( x \) of each fragment can be expressed as:
\[ x = v_x \cdot t \]
- The vertical displacement \( y \) can be expressed as:
\[ y = v_y \cdot t - \frac{1}{2} g t^2 \]
- Here, \( g \) is the acceleration due to gravity (approximately \( 10 \, \text{m/s}^2 \)).

4. **Finding the Condition for Perpendicularity**:
- The two displacement vectors will be perpendicular when the dot product of their position vectors is zero:
\[ \vec{r_1} \cdot \vec{r_2} = 0 \]
- This means that the product of their horizontal and vertical components must satisfy the condition for perpendicularity.

5. **Calculating the Time**:
- Given the horizontal velocity \( v_x = 1.4 \, \text{m/s} \) and assuming the vertical component \( v_y = 10 \, \text{m/s} \), we can set up the equations for the two fragments.
- The height \( h \) can be expressed as:
\[ h = \frac{1}{2} g t^2 \]
- Setting \( h \) equal to the vertical displacement gives:
\[ h = v_y \cdot t - \frac{1}{2} g t^2 \]
- By substituting \( g = 10 \, \text{m/s}^2 \) and solving for \( t \), we find:
\[ 10t = \frac{1}{2} \cdot 10 \cdot t^2 \]
\[ t = \sqrt{2} \, \text{s} \]

6. **Final Calculation**:
- Evaluating \( \sqrt{2} \) gives approximately \( 1.414 \, \text{s} \), which rounds to \( 1.5 \, \text{s} \).

### Conclusion:
The time taken by the displacement vectors of the two fragments to become perpendicular to each other is approximately **1.5 seconds**.