A bullet ofmass m is fired with a velocity `10m//s` at angle `theta` with horizontal. At the hightest point of its trajectory, it collsides head-on with a bob of mass `3m` suspended by a massless string of length `2//5m` and gets embedded in the bob. After th collision the string moves through an angle of `60^(@)`.
The vertical coordinate of the initial positon of the bob w.r.t. the point of firing of the bullet is

Options

A. `(9)/(4)m`
B. `(9)/(5)m`
C. `(24)/(5)m`
D. None of these

Explanation

To solve the problem step by step, we will follow these main points:

1. **Understanding the Problem**: A bullet of mass \( m \) is fired at an angle \( \theta \) with a velocity of \( 10 \, \text{m/s} \). At the highest point of its trajectory, it collides with a bob of mass \( 3m \) suspended by a massless string of length \( \frac{2}{5} \, \text{m} \). After the collision, the string moves through an angle of \( 60^\circ \). We need to find the vertical coordinate of the initial position of the bob with respect to the point of firing of the bullet.

2. **Finding the Horizontal Velocity of the Bullet**: At the highest point of the trajectory, the vertical component of the bullet's velocity is zero. The horizontal component of the velocity remains \( 10 \cos \theta \).

3. **Conservation of Momentum**: Before the collision, the momentum of the bullet is:
\[ p_{\text{initial}} = m \cdot (10 \cos \theta) + 0 = m \cdot (10 \cos \theta) \]
After the collision, the bullet and bob move together with a combined mass of \( 4m \) and a common velocity \( V \):
\[ p_{\text{final}} = 4m \cdot V \]
By conservation of momentum:
\[ m \cdot (10 \cos \theta) = 4m \cdot V \]
Simplifying gives:
\[ V = \frac{10 \cos \theta}{4} = \frac{5 \cos \theta}{2} \]

4. **Using Conservation of Energy**: The kinetic energy just after the collision is converted into potential energy as the bob rises. The change in kinetic energy is:
\[ \Delta KE = \frac{1}{2} \cdot 4m \cdot V^2 - 0 \]
The potential energy gained is:
\[ PE = 4mgh \]
Setting these equal gives:
\[ \frac{1}{2} \cdot 4m \cdot V^2 = 4mgh \]
Simplifying:
\[ \frac{1}{2} V^2 = gh \]

5. **Finding Height \( h \)**: We can express \( h \) in terms of \( V \):
\[ h = \frac{V^2}{2g} \]
Substituting \( V = \frac{5 \cos \theta}{2} \):
\[ h = \frac{(\frac{5 \cos \theta}{2})^2}{2g} = \frac{25 \cos^2 \theta}{8g} \]

6. **Using the Geometry of the Problem**: The bob moves through an angle of \( 60^\circ \). The vertical component of the string length gives:
\[ h = l(1 - \cos 60^\circ) = \frac{2}{5}(1 - \frac{1}{2}) = \frac{2}{5} \cdot \frac{1}{2} = \frac{1}{5} \, \text{m} \]

7. **Finding the Maximum Height of the Bullet**: The maximum height \( H \) attained by the bullet is given by:
\[ H = \frac{u^2 \sin^2 \theta}{2g} \]
Here, \( u = 10 \, \text{m/s} \) and \( g = 10 \, \text{m/s}^2 \):
\[ H = \frac{10^2 \sin^2 \theta}{2 \cdot 10} = \frac{100 \sin^2 \theta}{20} = 5 \sin^2 \theta \]

8. **Finding \( \sin^2 \theta \)**: From the triangle formed by \( \sin \theta \) and \( \cos \theta \):
\[ \sin^2 \theta + \cos^2 \theta = 1 \]
If \( \cos \theta = \frac{4}{5} \), then:
\[ \sin^2 \theta = 1 - \left(\frac{4}{5}\right)^2 = 1 - \frac{16}{25} = \frac{9}{25} \]

9. **Calculating \( H \)**:
\[ H = 5 \cdot \frac{9}{25} = \frac{45}{25} = 1.8 \, \text{m} \]

Thus, the vertical coordinate of the initial position of the bob with respect to the point of firing of the bullet is \( 1.8 \, \text{m} \).