A projectile of mass 3m is projected from ground with velocity `20sqrt2 m//s` at `45^@`. At highest point it explodes into two pieces. One of mass 2m and the other of mass m. Both the pieces fly off horizontally in opposite directions. Mass 2m falls at a distance of `100m` from point of projection. Find the distance of second mass from point of projection where it strikes the ground.
`(g=10m//s^2)`

Explanation

To solve the problem step by step, we will analyze the motion of the projectile, the explosion, and the subsequent motion of the two pieces.

### Step 1: Calculate the Range of the Projectile
The range \( R \) of a projectile is given by the formula:
\[ R = \frac{u^2 \sin(2\theta)}{g} \]
where:
- \( u = 20\sqrt{2} \, \text{m/s} \) (initial velocity)
- \( \theta = 45^\circ \) (angle of projection)
- \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity)

Calculating \( R \):
\[ R = \frac{(20\sqrt{2})^2 \sin(90^\circ)}{10} = \frac{800}{10} = 80 \, \text{m} \]

### Step 2: Analyze the Explosion
At the highest point of the projectile's trajectory, it explodes into two pieces:
- Mass \( 2m \) falls horizontally and lands \( 100 \, \text{m} \) away from the point of projection.
- Mass \( m \) will fall in the opposite direction.

### Step 3: Use the Center of Mass Concept
The center of mass of the system before and after the explosion remains unchanged. The position of the center of mass \( x_{cm} \) can be calculated using:
\[ x_{cm} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2} \]
where:
- \( m_1 = m \) (mass of the second piece)
- \( x_1 \) = distance where mass \( m \) lands (unknown)
- \( m_2 = 2m \) (mass of the first piece)
- \( x_2 = 100 \, \text{m} \) (distance where mass \( 2m \) lands)

Since the total mass is \( 3m \) and the center of mass position before the explosion is \( 80 \, \text{m} \):
\[ 80 = \frac{m \cdot x_1 + 2m \cdot 100}{3m} \]

### Step 4: Simplify the Equation
Multiplying through by \( 3m \) to eliminate the denominator:
\[ 240m = m \cdot x_1 + 200m \]
Dividing through by \( m \) (assuming \( m \neq 0 \)):
\[ 240 = x_1 + 200 \]

### Step 5: Solve for \( x_1 \)
Rearranging the equation gives:
\[ x_1 = 240 - 200 = 40 \, \text{m} \]

### Conclusion
The distance of the second mass \( m \) from the point of projection where it strikes the ground is \( 40 \, \text{m} \).