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 solve the problem, we will use the principles of conservation of momentum, kinematics, and projectile motion.

(a) Find the speeds of the 0.7 kg and 0.3 kg pieces immediately after the break-up.

Before the break-up at the highest point:

1. The horizontal and vertical components of the initial velocity of the projectile need to be determined.

\[ v_{x} = v_i \cdot \cos(\theta) = 60.0 \cdot \cos(41.0^\circ) \approx 60.0 \cdot 0.7547 \approx 45.3 , \text{m/s} \] \[ v_{y} = v_i \cdot \sin(\theta) = 60.0 \cdot \sin(41.0^\circ) \approx 60.0 \cdot 0.6561 \approx 39.4 , \text{m/s} \]

At the highest point, the vertical component of velocity is 0 m/s.

The total momentum before the break-up (only horizontal momentum since vertical momentum is zero at the highest point): \[ p_{initial} = m_{total} \cdot v_x = (2.0 , \text{kg}) \cdot (45.3 , \text{m/s}) = 90.6 , \text{kg·m/s} \]

After the break-up, the three parts must satisfy the conservation of momentum in both the \(x\) and \(y\) directions.

Let \(v_{0.7}\) be the speed of the 0.7 kg part (moving horizontally) and \(v_{0.3}\) be the speed of the 0.3 kg part (moving vertically):

Using conservation of momentum in the horizontal direction: \[ m_{1.0} \cdot 0 + m_{0.7} \cdot v_{0.7} + m_{0.3} \cdot 0 = p_{initial} \] \[ 0 + 0.7 \cdot v_{0.7} + 0.3 \cdot 0 = 90.6 \] \[ 0.7 \cdot v_{0.7} = 90.6 \rightarrow v_{0.7} = \frac{90.6}{0.7} \approx 129.43 , \text{m/s} \]

Now, using conservation of momentum in the vertical direction: \[ m_{0.3} \cdot v_{0.3} + m_{1.0} \cdot (-13.0) = 0 \] \[ 0.3 \cdot v_{0.3} + 1.0 \cdot (-13.0) = 0 \] \[ 0.3 \cdot v_{0.3} = 13.0 \rightarrow v_{0.3} = \frac{13.0}{0.3} \approx 43.33 , \text{m/s} \]

Thus, the speeds of the two parts immediately after the break-up are approximately:

• Speed of the 0.7 kg piece (\(v_{0.7}\)): 129.43 m/s (horizontal direction)
• Speed of the 0.3 kg piece (\(v_{0.3}\)): 43.33 m/s (vertical direction)

(b) How high does the 0.3 kg piece go before coming to rest?

To find the height, we can use the kinematic equation: \[ v_f^2 = v_i^2 + 2a d \] Where:

• \(v_f = 0\) m/s (final velocity at the highest point)
• \(v_i = 43.33\) m/s (initial velocity of the 0.3 kg piece)
• \(a = -g = -9.81 , \text{m/s}^2\) (acceleration due to gravity, acting downwards)
• \(d\) = height gained by the 0.3 kg piece

Plugging in these values: \[ 0 = (43.33)^2 + 2(-9.81)d \] \[ 0 = 1874.0289 - 19.62d \] \[ 19.62d = 1874.0289 \rightarrow d = \frac{1874.0289}{19.62} \approx 95.55 , \text{m} \]

Thus, the height the 0.3 kg piece goes before coming to rest is approximately:

• Height: 95.55 m

(c) Where does the 0.7 kg piece land relative to where it was fired from?

The 0.7 kg piece continues to move horizontally at the speed found earlier (\(v_{0.7} = 129.43 , \text{m/s}\)) with no vertical motion after the projectile breaks up.

To find the total time it is in the air, we first need to determine how long it takes for the 2.0 kg projectile to reach its highest point.

Using the vertical component of velocity: \[ t = \frac{v_{y}}{g} = \frac{39.4}{9.81} \approx 4.02 , \text{s} \]

This is the time to reach the highest point. The total time of flight to come back down is double this (up and down): \[ t_{total} = 2 \cdot 4.02 \approx 8.04 , \text{s} \]

Now, the horizontal distance traveled by the 0.7 kg piece while it is in the air: \[ d = v_{0.7} \cdot t_{total} = 129.43 , \text{m/s} \cdot 8.04 , \text{s} \approx 1040.9 , \text{m} \]

Thus, the horizontal distance from the launch point where the 0.7 kg piece lands is approximately:

• Horizontal distance: 1040.9 meters

This completes the answers to the questions posed.