A particle of mass `2m` is projected at an angle of `45^@` with horizontal with a velocity of `20sqrt2m//s`. After `1s` explosion takes place and the particle is broken into two equal pieces. As a result of explosion one part comes to rest. Find the maximum height attained by the other part. Take `g=10m//s^2`.

Explanation

To solve the problem step by step, we will follow the trajectory of the particle before and after the explosion, applying the principles of projectile motion and conservation of momentum.

### Step 1: Analyze the initial conditions
The particle of mass `2m` is projected at an angle of `45°` with a velocity of `20√2 m/s`. We need to find the initial horizontal (vx) and vertical (vy) components of the velocity.

**Calculation:**
- Horizontal component, \( v_{x} = v \cdot \cos(45°) = 20\sqrt{2} \cdot \frac{1}{\sqrt{2}} = 20 \, m/s \)
- Vertical component, \( v_{y} = v \cdot \sin(45°) = 20\sqrt{2} \cdot \frac{1}{\sqrt{2}} = 20 \, m/s \)

### Step 2: Determine the position of the particle after 1 second
After 1 second, we can find the vertical position (height) of the particle using the formula for vertical motion.

**Calculation:**
- The vertical position after 1 second can be calculated using the formula:
\[ h = v_{y} \cdot t - \frac{1}{2} g t^2 \]
Substituting \( v_{y} = 20 \, m/s \), \( g = 10 \, m/s^2 \), and \( t = 1 \, s \):
\[ h = 20 \cdot 1 - \frac{1}{2} \cdot 10 \cdot (1)^2 = 20 - 5 = 15 \, m \]

### Step 3: Apply conservation of momentum during the explosion
After 1 second, the particle explodes into two equal masses of `m`. One part comes to rest, and we need to find the velocity of the other part after the explosion.

**Calculation:**
- Let the velocity of the moving part after the explosion be \( v' \).
- By conservation of momentum:
\[ \text{Initial momentum} = \text{Final momentum} \]
Initial momentum (just before explosion):
\[ p_{initial} = 2m \cdot (20i + 20j) = 40mi + 40mj \]
Final momentum (after explosion):
\[ p_{final} = m \cdot 0 + m \cdot v' = mv' \]
Since one part is at rest, we can write:
\[ 40mi + 40mj = mv' \implies v' = (40i + 40j) \]

### Step 4: Find the vertical component of the velocity after the explosion
The vertical component of the velocity of the moving part after the explosion is \( v'_{y} = 20 \, m/s \).

### Step 5: Calculate the maximum height attained by the moving part
To find the maximum height attained by the moving part, we can use the formula for maximum height in projectile motion:
\[ H = \frac{v'_{y}^2}{2g} \]
Substituting \( v'_{y} = 20 \, m/s \) and \( g = 10 \, m/s^2 \):
\[ H = \frac{(20)^2}{2 \cdot 10} = \frac{400}{20} = 20 \, m \]

### Final Answer
The maximum height attained by the other part after the explosion is **20 meters**.

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