To solve the problem, we will use the principle of conservation of momentum. The projectile breaks into two equal parts at its highest point, and we need to find the velocity of the second part after the explosion.
### Step-by-Step Solution:
1. **Identify Initial Conditions**:
- The initial velocity of the projectile at its highest point is \( u = 20 \, \text{m/s} \) (horizontal direction).
- The mass of the projectile is \( 2m \) (since it breaks into two equal parts, each part has mass \( m \)).
2. **Momentum Before the Explosion**:
- The total initial momentum of the system (before the explosion) is:
\[
\text{Initial Momentum} = \text{mass} \times \text{velocity} = (2m) \times (20 \, \text{m/s}) = 40m \, \text{i}
\]
3. **Post-Explosion Conditions**:
- After the explosion, one part (let's call it Part A) moves vertically upwards with a velocity of \( 30 \, \text{m/s} \) (in the \( j \) direction).
- The momentum of Part A in the vertical direction is:
\[
\text{Momentum of A} = m \times (30 \, \text{m/s}) = 30m \, \text{j}
\]
4. **Momentum Conservation in the X-Direction**:
- Since there is no external force acting in the horizontal direction, the momentum in the x-direction before and after the explosion must be equal.
- Before the explosion: \( 40m \, \text{i} \)
- After the explosion, Part A has no x-component of momentum:
\[
40m \, \text{i} = m \cdot v_{B_x} \, \text{i}
\]
- Solving for \( v_{B_x} \):
\[
v_{B_x} = 40 \, \text{m/s}
\]
5. **Momentum Conservation in the Y-Direction**:
- The total initial momentum in the y-direction is zero (since the projectile is at its highest point).
- After the explosion:
\[
0 = 30m \, \text{j} + m \cdot v_{B_y} \, \text{j}
\]
- Solving for \( v_{B_y} \):
\[
v_{B_y} = -30 \, \text{m/s}
\]
6. **Finding the Velocity of Part B**:
- The velocity vector of Part B is:
\[
\vec{v}_B = v_{B_x} \, \text{i} + v_{B_y} \, \text{j} = 40 \, \text{i} - 30 \, \text{j} \, \text{m/s}
\]
7. **Magnitude of Velocity of Part B**:
- To find the magnitude of the velocity:
\[
|\vec{v}_B| = \sqrt{(40)^2 + (-30)^2} = \sqrt{1600 + 900} = \sqrt{2500} = 50 \, \text{m/s}
\]
### Final Answer:
The other part (Part B) will move with a velocity of \( 50 \, \text{m/s} \).
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