A projectile of mass 2.0 kg is fired in the air at an angle of 41.0° to the horizon at a speed of 60.0 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 13.0 m/s, the 0.7 kg part moves in the original forward direction, and the 0.3 kg part goes straight up.

To find the horizontal distance where the 0.7 kg piece lands after the breakup, we will first determine the time it remains in the air after the breakup and then multiply that time by its horizontal speed (velocity).

Step 1: Analyze the motion of the 0.7 kg piece.

1. Determine the horizontal and vertical components of the velocity of the original projectile at the peak of its trajectory:

   • The horizontal component (\(v_{x}\)) can be calculated as: \[ v_{x} = v \cdot \cos(\theta) = 60.0 , \text{m/s} \cdot \cos(41.0°) \]
   • The vertical component (\(v_{y}\)) is \(0\) at the highest point: \[ v_{y} = v \cdot \sin(\theta) = 60.0 , \text{m/s} \cdot \sin(41.0°) \]

   Calculation:

   • \(v_{x} = 60.0 \cdot \cos(41.0°) \approx 60.0 \cdot 0.7547 \approx 452.82 , \text{m/s}\)
   • \(v_{y} = 60.0 \cdot \sin(41.0°) \approx 60.0 \cdot 0.6561 \approx 39.37 , \text{m/s}\)

   Therefore, the horizontal velocity of the 0.7 kg piece after the breakup remains unchanged at \(v_{x} = 452.82 , \text{m/s}\) (as it continues in the same horizontal direction) since no external horizontal forces act on it in the horizontal direction.

Step 2: Calculate the time the 0.7 kg piece is in the air after the breakup.

2. We need to calculate the time it takes for the 0.3 kg piece to reach a maximum height and then fall back to the original height (the height at which the breakup occurred).

Using the kinematic equation for vertical motion: \[ v_{f} = v_{i} - g t \]

At maximum height, \(v_{f} = 0\), so: \[ 0 = v_{3} - g t_{up} \] \[ t_{up} = \frac{v_{3}}{g} = \frac{43.3 , \text{m/s}}{9.81 , \text{m/s}^2} \approx 4.41 , \text{s} \]

The time to fall back down is equal to the time to go up, so total time \(t_{total} = 2t_{up}\): \[ t_{total} = 2 \cdot 4.41 \approx 8.82 , \text{s} \]

Step 3: Calculate the horizontal distance.

3. Use the horizontal velocity and total time to find the horizontal distance: \[ d = v_{x} \cdot t_{total} \approx 452.82 , \text{m/s} \cdot 8.82 , \text{s} \approx 3999.74 , \text{m} \]

Answer: The 0.7 kg piece lands approximately \(3999.74 , \text{m}\) from where it was fired from.