At a certain height a shell at rest explodes into two equal fragments. One of the fragments receives a horizontal velocity u. The time interval after which, the velocity vectors will be inclined at `120^(@)` to each other is

Options

A. `(u)/(sqrt3g)`
B. `(sqrt3u)/(g)`
C. `(2u)/(sqrt3g)`
D. `(u)/(2sqrt3g)`

Answer

To solve the problem step by step, we will follow the principles of conservation of momentum and kinematics.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- A shell at rest explodes into two equal fragments.
- One fragment receives a horizontal velocity \( u \).
- We need to find the time interval after which the velocity vectors of the two fragments are inclined at an angle of \( 120^\circ \).

2. **Initial Conditions**:
- Before the explosion, the total momentum of the system is zero since the shell is at rest.
- After the explosion, let the first fragment (Fragment 1) have a velocity \( \vec{v_1} = u \hat{i} + gt \hat{j} \) (moving horizontally with velocity \( u \) and falling under gravity).
- Let the second fragment (Fragment 2) have a velocity \( \vec{v_2} = -u \hat{i} + gt \hat{j} \) (moving in the opposite horizontal direction and also falling under gravity).

3. **Using the Cosine Rule**:
- The angle between the two velocity vectors is given by the formula:
\[ \cos \theta = \frac{\vec{v_1} \cdot \vec{v_2}}{|\vec{v_1}| |\vec{v_2}|} \]
- We know that \( \theta = 120^\circ \), thus \( \cos 120^\circ = -\frac{1}{2} \).

4. **Calculating the Dot Product**:
- Calculate \( \vec{v_1} \cdot \vec{v_2} \):
\[ \vec{v_1} \cdot \vec{v_2} = (u \hat{i} + gt \hat{j}) \cdot (-u \hat{i} + gt \hat{j}) = -u^2 + g^2 t^2 \]

5. **Calculating the Magnitudes**:
- The magnitude of \( \vec{v_1} \) and \( \vec{v_2} \):
\[ |\vec{v_1}| = |\vec{v_2}| = \sqrt{u^2 + (gt)^2} \]

6. **Setting Up the Equation**:
- Substitute into the cosine formula:
\[ -\frac{1}{2} = \frac{-u^2 + g^2 t^2}{u^2 + g^2 t^2} \]

7. **Cross Multiplying**:
- Rearranging gives:
\[ -\frac{1}{2} (u^2 + g^2 t^2) = -u^2 + g^2 t^2 \]
- This simplifies to:
\[ -\frac{1}{2} u^2 - \frac{1}{2} g^2 t^2 = -u^2 + g^2 t^2 \]

8. **Combining Like Terms**:
- Rearranging terms:
\[ \frac{1}{2} u^2 = \frac{3}{2} g^2 t^2 \]

9. **Solving for Time \( t \)**:
- Rearranging gives:
\[ t^2 = \frac{u^2}{3g^2} \]
- Taking the square root:
\[ t = \frac{u}{\sqrt{3}g} \]

### Final Answer:
Thus, the time interval after which the velocity vectors will be inclined at \( 120^\circ \) to each other is:
\[ t = \frac{u}{\sqrt{3}g} \]