A projectile of mass 2.0 kg is fired in the air at an angle of 36.0° to the horizon at a speed of 44.0 m/s. At the highest point in its flight, the projectile breaks into three parts of mass 1.0 kg, 0.7 kg, and 0.3 kg. The 1.0 kg part falls straight down after breakup with an initial speed of 10.4 m/s, the 0.7 kg part moves in the original forward direction, and the 0.3 kg part goes straight up. At peak after explosion V3,1 = (V3,)} m3 = 0.3 kg At peak before explosion m2 = 0.7 kg V21 = (V2,)î Vx= (Vixî = (44.0 m/s)v, m, = 1.0 kg ,f = -(10.4 m/s)j 36.0° Launch (a) Find the speeds (in m/s) of the 0.7 kg and 0.3 kg pieces immediately after the break-up. Vっ = 101.7 m/s V3 = 34.67 m/s (b) How high (in m) from the break-up point does the 0.3 kg piece go before coming to rest? 61.32 (c) Where does the 0.7 kg piece land relative to where it was fired from? (Give the horizontal distance in meters from the launch point to where the 0.7 kg piece lands.) 546 X m

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How would I go about solving part c?

**Projectile Motion and Breakup Analysis**

A projectile of mass 2.0 kg is fired in the air at an angle of 36.0° to the horizon at a speed of 44.0 m/s. At the highest point in its flight, the projectile breaks into three parts of mass 1.0 kg, 0.7 kg, and 0.3 kg. The 1.0 kg part falls straight down after breakup with an initial speed of 10.4 m/s, the 0.7 kg part moves in the original forward direction, and the 0.3 kg part goes straight up.

### Diagram Explanation

- **Launch:** 
  - Initial velocity vector (\(\vec{v}_i = (44.0 \, \text{m/s})\hat{v}_i\)) is depicted launching upwards at a 36.0° angle.

- **At Peak Before Explosion:**
  - The projectile is at its highest point before breaking up.

- **At Peak After Explosion:**
  - **\(m_1 = 1.0\, \text{kg}\):** Moves straight down with \(\vec{v}_{1,f} = (-10.4 \, \text{m/s})\hat{s}\).
  - **\(m_2 = 0.7\, \text{kg}\):** Moves forward with velocity \(\vec{v}_{2,f} = (v_{2,t})\hat{i}\).
  - **\(m_3 = 0.3\, \text{kg}\):** Moves upwards with velocity \(\vec{v}_{3,f} = (v_{3})\hat{j}\).

### Calculations

(a) **Find the speeds (in m/s) of the 0.7 kg and 0.3 kg pieces immediately after the break-up.**
- \(v_2 = 101.7 \, \text{m/s}\) 
- \(v_3 = 34.67 \, \text{m/s}\)

(b) **How high (in m) from the break-up point does the 0.3 kg piece go before coming to rest?**
- Height = 61.32 m

(c) **Where does the 0.7 kg piece land relative to where it was fired from? (Give the horizontal distance in meters from the launch point
Transcribed Image Text:**Projectile Motion and Breakup Analysis** A projectile of mass 2.0 kg is fired in the air at an angle of 36.0° to the horizon at a speed of 44.0 m/s. At the highest point in its flight, the projectile breaks into three parts of mass 1.0 kg, 0.7 kg, and 0.3 kg. The 1.0 kg part falls straight down after breakup with an initial speed of 10.4 m/s, the 0.7 kg part moves in the original forward direction, and the 0.3 kg part goes straight up. ### Diagram Explanation - **Launch:** - Initial velocity vector (\(\vec{v}_i = (44.0 \, \text{m/s})\hat{v}_i\)) is depicted launching upwards at a 36.0° angle. - **At Peak Before Explosion:** - The projectile is at its highest point before breaking up. - **At Peak After Explosion:** - **\(m_1 = 1.0\, \text{kg}\):** Moves straight down with \(\vec{v}_{1,f} = (-10.4 \, \text{m/s})\hat{s}\). - **\(m_2 = 0.7\, \text{kg}\):** Moves forward with velocity \(\vec{v}_{2,f} = (v_{2,t})\hat{i}\). - **\(m_3 = 0.3\, \text{kg}\):** Moves upwards with velocity \(\vec{v}_{3,f} = (v_{3})\hat{j}\). ### Calculations (a) **Find the speeds (in m/s) of the 0.7 kg and 0.3 kg pieces immediately after the break-up.** - \(v_2 = 101.7 \, \text{m/s}\) - \(v_3 = 34.67 \, \text{m/s}\) (b) **How high (in m) from the break-up point does the 0.3 kg piece go before coming to rest?** - Height = 61.32 m (c) **Where does the 0.7 kg piece land relative to where it was fired from? (Give the horizontal distance in meters from the launch point
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