A projectile of mass `3kg` is projected with velocity `50m//s` at `37^@` from horizontal. After `2s`, explosion takes place and the projectile breaks into two parts of masses `1kg` and `2kg`. The first part comes to rest just after explosion.
Find,
(a) the velocity of second part just after explosion.
(b) maximum height attained by this part. Take `g=10m//s^2`

Explanation

To solve the problem step by step, we will break it down into two parts as requested: (a) finding the velocity of the second part just after the explosion, and (b) calculating the maximum height attained by this part.

### Step-by-Step Solution

**Given Data:**
- Mass of the projectile (m) = 3 kg
- Initial velocity (u) = 50 m/s
- Angle of projection (θ) = 37°
- Time before explosion (t) = 2 s
- Mass of the first part after explosion (m1) = 1 kg
- Mass of the second part after explosion (m2) = 2 kg
- Acceleration due to gravity (g) = 10 m/s²

### Part (a): Velocity of the second part just after the explosion

1. **Resolve the initial velocity into components:**
- Horizontal component (u_x) = u * cos(θ) = 50 * cos(37°) = 50 * (4/5) = 40 m/s
- Vertical component (u_y) = u * sin(θ) = 50 * sin(37°) = 50 * (3/5) = 30 m/s

2. **Calculate the velocity after 2 seconds:**
- The horizontal velocity remains constant: \( v_x = u_x = 40 \, \text{m/s} \)
- The vertical velocity after 2 seconds (using \( v_y = u_y - g \cdot t \)):
\[ v_y = 30 - 10 \cdot 2 = 30 - 20 = 10 \, \text{m/s} \]
- Therefore, the velocity vector just before the explosion is:
\[ \vec{v} = 40 \hat{i} + 10 \hat{j} \, \text{m/s} \]

3. **Apply conservation of momentum:**
- Before the explosion, the total momentum is:
\[ \text{Total momentum} = m \cdot \vec{v} = 3 \cdot (40 \hat{i} + 10 \hat{j}) = 120 \hat{i} + 30 \hat{j} \, \text{kg m/s} \]
- After the explosion, the first part (1 kg) comes to rest, so its momentum is:
\[ \text{Momentum of first part} = 1 \cdot 0 = 0 \, \text{kg m/s} \]
- Let the velocity of the second part (2 kg) be \( \vec{v_2} \). Therefore, we have:
\[ 120 \hat{i} + 30 \hat{j} = 0 + 2 \cdot \vec{v_2} \]
- Solving for \( \vec{v_2} \):
\[ \vec{v_2} = \frac{120 \hat{i} + 30 \hat{j}}{2} = 60 \hat{i} + 15 \hat{j} \, \text{m/s} \]

**Answer for Part (a):** The velocity of the second part just after the explosion is \( 60 \hat{i} + 15 \hat{j} \, \text{m/s} \).

### Part (b): Maximum height attained by the second part

1. **Calculate the height at which the explosion occurs:**
- Use the vertical displacement formula:
\[ h = u_y \cdot t - \frac{1}{2} g t^2 \]
\[ h = 30 \cdot 2 - \frac{1}{2} \cdot 10 \cdot (2^2) = 60 - 20 = 40 \, \text{m} \]

2. **Calculate the additional height gained by the second part after the explosion:**
- The vertical component of the velocity of the second part is \( 15 \, \text{m/s} \).
- Use the formula for maximum height gained after the explosion:
\[ h_{additional} = \frac{v_y^2}{2g} = \frac{15^2}{2 \cdot 10} = \frac{225}{20} = 11.25 \, \text{m} \]

3. **Total maximum height attained by the second part:**
- Total height = height at explosion + additional height:
\[ \text{Total height} = 40 + 11.25 = 51.25 \, \text{m} \]

**Answer for Part (b):** The maximum height attained by the second part is \( 51.25 \, \text{m} \).