A projectile of mass 3m explodes at highest point of its path. It breaks into three equalparts. One part retraces its path, the second one comes to rest. The range of the projectile was 100 m if no explosion would have taken place. The distance of the third part from the point of projection when it finally lands on the ground is -

Options

A. 100m
B. 150m
C. 250m
D. 300m

Answer

To solve the problem, we need to analyze the motion of the projectile before and after the explosion.

### Step-by-Step Solution:

1. **Understanding the Initial Conditions**:
- The projectile has a mass of \(3m\) and reaches a maximum height before exploding. The range of the projectile without the explosion is given as \(100 \, \text{m}\).

2. **Explosion at the Highest Point**:
- At the highest point, the vertical component of the velocity is \(0\), and the horizontal component of the velocity remains constant. Let the horizontal velocity of the projectile be \(v_x\).

3. **Calculating the Horizontal Velocity**:
- The range \(R\) of a projectile is given by the formula:
\[ R = \frac{v^2 \sin(2\theta)}{g} \]
- Here, \(R = 100 \, \text{m}\). If we assume the angle of projection is \(\theta\), we can express the horizontal component of the velocity as:
\[ v_x = v \cos(\theta) \]
- The time of flight \(T\) can be derived from the range formula:
\[ T = \frac{R}{v_x} = \frac{100}{v \cos(\theta)} \]

4. **After the Explosion**:
- The projectile explodes into three equal parts, each of mass \(m\).
- One part retraces its path, meaning it moves back with the same horizontal velocity \(v_x\).
- The second part comes to rest, meaning it does not move at all.
- The third part's motion needs to be analyzed.

5. **Momentum Conservation**:
- Since the explosion occurs at the highest point, the horizontal momentum before and after the explosion must be conserved.
- Before the explosion, the total horizontal momentum is:
\[ P_{\text{initial}} = 3m \cdot v_x \]
- After the explosion, the momentum is:
\[ P_{\text{final}} = m \cdot (-v_x) + m \cdot 0 + m \cdot v_{3x} \]
- Setting these equal gives:
\[ 3m \cdot v_x = -m \cdot v_x + m \cdot v_{3x} \]
- Simplifying, we find:
\[ 4v_x = v_{3x} \quad \Rightarrow \quad v_{3x} = 4v_x \]

6. **Finding the Distance Traveled by the Third Part**:
- The third part moves horizontally with a velocity of \(4v_x\) and falls from the highest point. The time taken to fall to the ground is the same as the time of flight of the original projectile, which is \(T\).
- The horizontal distance traveled by the third part is:
\[ d_3 = v_{3x} \cdot T = 4v_x \cdot T \]
- Substituting \(T = \frac{100}{v_x}\):
\[ d_3 = 4v_x \cdot \frac{100}{v_x} = 400 \, \text{m} \]

### Final Answer:
The distance of the third part from the point of projection when it finally lands on the ground is **400 m**.