A rocket of mass `20kg` has `180kg` fuel. The exhaust velocity of the fuel is `1.6km//s`. Calculate the minimum rate of consumption of fuel so that the rocket may rise from the ground. Also, calculate the ultimate vertical speed gained by the rocket when the rate of consumption of fuel is `(g=9.8m//s^2)`
(i) `2kg//s` (ii) `20kg//s`

Explanation

To solve the problem, we will break it down into two parts:

1. **Calculating the minimum rate of fuel consumption for the rocket to rise from the ground.**
2. **Calculating the ultimate vertical speed gained by the rocket for two different rates of fuel consumption (2 kg/s and 20 kg/s).**

### Part 1: Minimum Rate of Fuel Consumption

1. **Identify the given values:**
- Mass of the rocket (m₀) = 20 kg
- Mass of the fuel (m_fuel) = 180 kg
- Total initial mass (m_total) = m₀ + m_fuel = 20 kg + 180 kg = 200 kg
- Exhaust velocity (v_e) = 1.6 km/s = 1600 m/s
- Acceleration due to gravity (g) = 9.8 m/s²

2. **Determine the condition for the rocket to just rise:**
- For the rocket to rise, the thrust (F_t) must equal the weight (W) of the rocket.
- Thrust can be expressed as:
\[ F_t = -\frac{dm}{dt} \cdot v_e \]
- Weight of the rocket is given by:
\[ W = m_{total} \cdot g = 200 \cdot 9.8 \]

3. **Set the thrust equal to the weight:**
\[ -\frac{dm}{dt} \cdot v_e = m_{total} \cdot g \]
Rearranging gives:
\[ \frac{dm}{dt} = -\frac{m_{total} \cdot g}{v_e} \]

4. **Substituting the values:**
\[ \frac{dm}{dt} = -\frac{200 \cdot 9.8}{1600} \]

5. **Calculating the minimum rate of fuel consumption:**
\[ \frac{dm}{dt} = -\frac{1960}{1600} = -1.225 \text{ kg/s} \]
Since we are interested in the rate of fuel consumption, we take the positive value:
\[ \frac{dm}{dt} = 1.225 \text{ kg/s} \]

### Part 2: Ultimate Vertical Speed Gained by the Rocket

1. **For a rate of fuel consumption of 2 kg/s:**
- Time to consume all fuel:
\[ t = \frac{m_fuel}{\frac{dm}{dt}} = \frac{180}{2} = 90 \text{ seconds} \]
- Final mass of the rocket after fuel is consumed:
\[ m = m₀ = 20 \text{ kg} \]
- Using the formula for final velocity:
\[ v = u - gt + v_e \ln\left(\frac{m_{total}}{m}\right) \]
- Substituting the values:
\[ v = 0 - (9.8 \cdot 90) + (1600) \ln\left(\frac{200}{20}\right) \]
- Calculate:
\[ v = -882 + 1600 \cdot \ln(10) \]
- Using \(\ln(10) \approx 2.302\):
\[ v \approx -882 + 1600 \cdot 2.302 \approx -882 + 3683.2 \approx 2801.2 \text{ m/s} \approx 2.8012 \text{ km/s} \]

2. **For a rate of fuel consumption of 20 kg/s:**
- Time to consume all fuel:
\[ t = \frac{180}{20} = 9 \text{ seconds} \]
- Final mass of the rocket:
\[ m = 20 \text{ kg} \]
- Using the same formula:
\[ v = 0 - (9.8 \cdot 9) + (1600) \ln\left(\frac{200}{20}\right) \]
- Calculate:
\[ v = -88.2 + 1600 \cdot \ln(10) \approx -88.2 + 3683.2 \approx 3595 \text{ m/s} \approx 3.595 \text{ km/s} \]

### Final Results:
- Minimum rate of fuel consumption: **1.225 kg/s**
- Ultimate vertical speed for 2 kg/s: **2.8012 km/s**
- Ultimate vertical speed for 20 kg/s: **3.595 km/s**