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

Explanation

To solve the problem, we need to find the height of the third part of the shell after it explodes. Let's break down the solution step by step.

### Step 1: Understand the scenario
The shell has a mass of 4 kg and is moving upward with a velocity of 10 m/s. It explodes at a height of 50 m. After 3 seconds, one part (2 kg) reaches the ground, another part (1 kg) is at a height of 40 m, and we need to find the height of the third part.

### Step 2: Calculate the height of the center of mass after the explosion
The center of mass (COM) of the system can be calculated using the formula:

\[ \text{COM} = \frac{m_1 h_1 + m_2 h_2 + m_3 h_3}{m_1 + m_2 + m_3} \]

Here, \(m_1 = 2 \, \text{kg}\) (on the ground, height = 0), \(m_2 = 1 \, \text{kg}\) (height = 40 m), and \(m_3 = 1 \, \text{kg}\) (height = \(y\)).

### Step 3: Calculate the height of the center of mass after 3 seconds
The height of the center of mass after 3 seconds can be calculated as follows:

1. The initial height of the shell before the explosion is 50 m.
2. The height after 3 seconds can be calculated using the equation of motion:

\[ h = h_0 + ut - \frac{1}{2}gt^2 \]

Where:
- \(h_0 = 50 \, \text{m}\)
- \(u = 10 \, \text{m/s}\)
- \(g = 10 \, \text{m/s}^2\)
- \(t = 3 \, \text{s}\)

Substituting the values:

\[ h = 50 + (10 \times 3) - \frac{1}{2} \times 10 \times (3^2) \]
\[ h = 50 + 30 - \frac{1}{2} \times 10 \times 9 \]
\[ h = 50 + 30 - 45 \]
\[ h = 35 \, \text{m} \]

So, the height of the center of mass after 3 seconds is 35 m.

### Step 4: Set up the equation for the center of mass
Now we can set up the equation for the center of mass:

\[ 35 = \frac{(2 \times 0) + (1 \times 40) + (1 \times y)}{2 + 1 + 1} \]

This simplifies to:

\[ 35 = \frac{0 + 40 + y}{4} \]

### Step 5: Solve for \(y\)
Now, multiplying both sides by 4:

\[ 140 = 40 + y \]

Now, solving for \(y\):

\[ y = 140 - 40 \]
\[ y = 100 \, \text{m} \]

### Conclusion
The height of the third part from the ground is **100 m**.