Question

physics projectile motion problems 1-2 km/s 50-60 degrees

Answer 1

To analyze projectile motion at initial speeds between 1-2 km/s and launch angles between 50 and 60 degrees, we can use the basic kinematic equations of projectile motion. This type of motion can be broken down into horizontal and vertical components, which can be analyzed separately.

### Key Parameters

1. **Initial Speed (v₀)**: This can vary from 1,000 m/s to 2,000 m/s (1-2 km/s).
2. **Launch Angle (θ)**: Between 50 and 60 degrees.
3. **Acceleration due to Gravity (g)**: Approximately \(9.81 \, \text{m/s}^2\).

### Components of Initial Velocity

The initial velocity can be broken into horizontal and vertical components:

- \( v_{0x} = v_0 \cdot \cos(\theta) \)
- \( v_{0y} = v_0 \cdot \sin(\theta) \)

### Time of Flight (T)

The time of flight until the projectile returns to the initial height can be determined using the formula:

\[ T = \frac{2v_{0y}}{g} \]

### Maximum Height (H)

The maximum height reached can be calculated using:

\[ H = \frac{v_{0y}^2}{2g} \]

### Range (R)

The horizontal range can be determined by:

\[ R = v_{0x} \cdot T = v_{0x} \cdot \frac{2v_{0y}}{g} \]

### Example Calculations

#### Case 1: v₀ = 1000 m/s, θ = 50°

1. Calculate components of initial velocity:
- \( v_{0x} = 1000 \cdot \cos(50°) \approx 643.0 \, \text{m/s} \)
- \( v_{0y} = 1000 \cdot \sin(50°) \approx 766.0 \, \text{m/s} \)

2. Time of Flight:
\[ T = \frac{2 \cdot 766}{9.81} \approx 156.4 \, \text{s} \]

3. Maximum Height:
\[ H = \frac{766^2}{2 \cdot 9.81} \approx 30,030.1 \, \text{m} \]

4. Range:
\[ R = 643.0 \cdot 156.4 \approx 100,628.5 \, \text{m} \text{ or } 100.6 \, \text{km} \]

#### Case 2: v₀ = 2000 m/s, θ = 60°

1. Calculate components of initial velocity:
- \( v_{0x} = 2000 \cdot \cos(60°) = 1000 \, \text{m/s} \)
- \( v_{0y} = 2000 \cdot \sin(60°) \approx 1732.1 \, \text{m/s} \)

2. Time of Flight:
\[ T = \frac{2 \cdot 1732.1}{9.81} \approx 352.4 \, \text{s} \]

3. Maximum Height:
\[ H = \frac{1732.1^2}{2 \cdot 9.81} \approx 28,022.1 \, \text{m} \]

4. Range:
\[ R = 1000 \cdot 352.4 \approx 352,400 \, \text{m} \text{ or } 352.4 \, \text{km} \]

### Summary of Results

- For \(v_0 = 1000 \, \text{m/s}, \theta = 50^\circ\):
- Time of Flight: ≈ 156.4 s
- Maximum Height: ≈ 30,030.1 m
- Range: ≈ 100.6 km

- For \(v_0 = 2000 \, \text{m/s}, \theta = 60^\circ\):
- Time of Flight: ≈ 352.4 s
- Maximum Height: ≈ 28,022.1 m
- Range: ≈ 352.4 km

These calculations illustrate the significant effects of both the initial speed and angle on the projectile's trajectory. The values can vary for different angles within the specified range, and it is possible to create similar calculations for any combination of the parameters given.