To solve the problem step-by-step, we will break it down into parts.
### Given Data:
- Mass of the rocket (m_rocket) = 20 kg
- Mass of the fuel (m_fuel) = 180 kg
- Total mass (m_0) = m_rocket + m_fuel = 20 kg + 180 kg = 200 kg
- Exhaust velocity (V_r) = 1.6 km/s = 1600 m/s
- Acceleration due to gravity (g) = 9.8 m/s²
### Part 1: Minimum Rate of Consumption of Fuel
1. **Calculate the weight of the rocket**:
\[
W = m_0 \cdot g = 200 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 = 1960 \, \text{N}
\]
2. **Establish the thrust force (F_t)**:
The thrust force must equal the weight of the rocket for it to just rise:
\[
F_t = W
\]
3. **Relate thrust force to rate of fuel consumption**:
The thrust force can also be expressed as:
\[
F_t = V_r \cdot \left(-\frac{dm}{dt}\right)
\]
where \(-\frac{dm}{dt}\) is the rate of fuel consumption.
4. **Set the equations equal**:
\[
V_r \cdot \left(-\frac{dm}{dt}\right) = W
\]
Rearranging gives:
\[
-\frac{dm}{dt} = \frac{W}{V_r}
\]
5. **Substitute known values**:
\[
-\frac{dm}{dt} = \frac{1960 \, \text{N}}{1600 \, \text{m/s}} = 1.225 \, \text{kg/s}
\]
Therefore, the minimum rate of consumption of fuel is:
\[
\frac{dm}{dt} \approx 1.225 \, \text{kg/s}
\]
### Part 2: Ultimate Vertical Speed for Different Rates of Consumption
#### (i) For a rate of consumption of fuel = 2 kg/s
1. **Calculate the time until fuel is exhausted**:
\[
T = \frac{m_fuel}{\text{Rate of consumption}} = \frac{180 \, \text{kg}}{2 \, \text{kg/s}} = 90 \, \text{s}
\]
2. **Calculate the final mass when fuel is exhausted**:
\[
m = m_0 - m_fuel = 200 \, \text{kg} - 180 \, \text{kg} = 20 \, \text{kg}
\]
3. **Use the formula for ultimate vertical speed**:
\[
V = u - gT + V_r \cdot \ln\left(\frac{m_0}{m}\right)
\]
where \(u = 0\) (initial velocity).
4. **Substituting the values**:
\[
V = 0 - 9.8 \cdot 90 + 1600 \cdot \ln\left(\frac{200}{20}\right)
\]
\[
V = -882 + 1600 \cdot \ln(10)
\]
\[
V \approx -882 + 1600 \cdot 2.302 \approx -882 + 3683.2 \approx 2801.2 \, \text{m/s}
\]
#### (ii) For a rate of consumption of fuel = 20 kg/s
1. **Calculate the time until fuel is exhausted**:
\[
T = \frac{180 \, \text{kg}}{20 \, \text{kg/s}} = 9 \, \text{s}
\]
2. **Calculate the final mass when fuel is exhausted**:
\[
m = 200 \, \text{kg} - 180 \, \text{kg} = 20 \, \text{kg}
\]
3. **Use the same formula for ultimate vertical speed**:
\[
V = 0 - 9.8 \cdot 9 + 1600 \cdot \ln\left(\frac{200}{20}\right)
\]
\[
V = -88.2 + 1600 \cdot 2.302 \approx -88.2 + 3683.2 \approx 3595 \, \text{m/s}
\]
### Final Answers:
1. Minimum rate of consumption of fuel: **1.225 kg/s**
2. Ultimate vertical speed for 2 kg/s: **2801.2 m/s**
3. Ultimate vertical speed for 20 kg/s: **3595 m/s**
