A shell of mass `4kg` moving with a velocoity `10m//s` vertically upward explodes into three parts at a height `50m` from ground. After three seconds, one part of mass `2kg` reaches ground and another part of mass `1kg` is at height `40m` from ground. The height of the third part from the ground is : `[g=10m//s^(2)]`

Options

A. 50 m
B. 80 m
C. 100 m
D. none of these

Answer

To solve the problem, we need to determine the height of the third part of the shell after it explodes. We will use the concept of the center of mass and the equations of motion.

### Step-by-Step Solution:

1. **Determine the initial conditions:**
- The mass of the shell, \( M = 4 \, \text{kg} \)
- The initial velocity, \( u = 10 \, \text{m/s} \)
- The height at which it explodes, \( h_0 = 50 \, \text{m} \)
- The time after the explosion, \( t = 3 \, \text{s} \)

2. **Calculate the height of the center of mass after 3 seconds:**
The height of the center of mass \( h_{cm} \) after time \( t \) can be calculated using the formula:
\[ h_{cm} = h_0 + ut - \frac{1}{2}gt^2 \]
Substituting the values:
\[ h_{cm} = 50 + (10 \times 3) - \frac{1}{2} \times 10 \times (3^2) \]
\[ h_{cm} = 50 + 30 - \frac{1}{2} \times 10 \times 9 \]
\[ h_{cm} = 50 + 30 - 45 = 35 \, \text{m} \]

3. **Set up the equation for the center of mass:**
The center of mass of the three parts can be expressed as:
\[ h_{cm} = \frac{m_1 h_1 + m_2 h_2 + m_3 h_3}{m_1 + m_2 + m_3} \]
Where:
- \( m_1 = 2 \, \text{kg} \) (reaches the ground, so \( h_1 = 0 \))
- \( m_2 = 1 \, \text{kg} \) (at height \( h_2 = 40 \, \text{m} \))
- \( m_3 = 1 \, \text{kg} \) (height \( h_3 \) is unknown)

Substituting the known values:
\[ 35 = \frac{(2 \times 0) + (1 \times 40) + (1 \times h_3)}{2 + 1 + 1} \]
\[ 35 = \frac{0 + 40 + h_3}{4} \]

4. **Solve for \( h_3 \):**
Multiply both sides by 4:
\[ 140 = 40 + h_3 \]
Rearranging gives:
\[ h_3 = 140 - 40 = 100 \, \text{m} \]

### Conclusion:
The height of the third part from the ground is \( h_3 = 100 \, \text{m} \).