(a) A rocket set for vertical firing weighs `50kg` and contains `450kg` of fuel. It can have a maximum exhaust velocity of `2km//s`. What should be its minimum rate of fuel consumption
(i) to just lift off the launching pad?
(ii) to give it an initial acceleration of `20m//s^2`?
(b) What will be the speed of the rocket when the rate of consumption of fuel is `10kg//s` after whole of the fuel is consumed? (Take `g=9.8m//s^2`)

Explanation

To solve the problem step by step, we will break it down into parts (a) and (b) as specified in the question.

### Part (a)

**(i) To just lift off the launching pad:**

1. **Identify the forces acting on the rocket:**
- The thrust force \( F_t \) must equal the weight of the rocket to just lift off.
- The weight \( W \) of the rocket can be calculated as:
\[ W = (m_{rocket} + m_{fuel}) \cdot g \]
where \( m_{rocket} = 50 \, \text{kg} \), \( m_{fuel} = 450 \, \text{kg} \), and \( g = 9.8 \, \text{m/s}^2 \).

2. **Calculate the total weight:**
\[ W = (50 + 450) \cdot 9.8 = 500 \cdot 9.8 = 4900 \, \text{N} \]

3. **Relate thrust force to fuel consumption:**
- The thrust force can also be expressed in terms of the rate of fuel consumption \( \frac{dm}{dt} \) and exhaust velocity \( v_e \):
\[ F_t = v_e \cdot \frac{dm}{dt} \]
where \( v_e = 2000 \, \text{m/s} \).

4. **Set the thrust force equal to the weight:**
\[ v_e \cdot \frac{dm}{dt} = W \]
\[ 2000 \cdot \frac{dm}{dt} = 4900 \]

5. **Solve for the rate of fuel consumption \( \frac{dm}{dt} \):**
\[ \frac{dm}{dt} = \frac{4900}{2000} = 2.45 \, \text{kg/s} \]

**(ii) To give it an initial acceleration of \( 20 \, \text{m/s}^2 \):**

1. **Apply Newton's second law:**
- The net force \( F_{net} \) acting on the rocket is given by:
\[ F_{net} = m \cdot a = (m_{rocket} + m_{fuel}) \cdot a \]
where \( a = 20 \, \text{m/s}^2 \).

2. **Calculate the net force required:**
\[ F_{net} = (50 + 450) \cdot 20 = 500 \cdot 20 = 10000 \, \text{N} \]

3. **Account for the weight of the rocket:**
- The thrust must overcome both the weight and provide the necessary acceleration:
\[ F_t = F_{net} + W \]
\[ F_t = 10000 + 4900 = 14900 \, \text{N} \]

4. **Relate thrust force to fuel consumption:**
\[ 2000 \cdot \frac{dm}{dt} = 14900 \]

5. **Solve for the rate of fuel consumption \( \frac{dm}{dt} \):**
\[ \frac{dm}{dt} = \frac{14900}{2000} = 7.45 \, \text{kg/s} \]

### Part (b)

**To find the speed of the rocket when the rate of consumption of fuel is \( 10 \, \text{kg/s} \) after all the fuel is consumed:**

1. **Calculate the time taken to consume all fuel:**
\[ \text{Time} = \frac{m_{fuel}}{\frac{dm}{dt}} = \frac{450}{10} = 45 \, \text{s} \]

2. **Use the rocket equation to find the final speed:**
- The final speed \( V \) can be calculated using the formula:
\[ V = V_e \cdot \ln\left(\frac{m_0}{m}\right) - g \cdot t \]
where:
- \( V_e = 2000 \, \text{m/s} \)
- \( m_0 = m_{rocket} + m_{fuel} = 500 \, \text{kg} \)
- \( m = m_{rocket} = 50 \, \text{kg} \)
- \( g = 9.8 \, \text{m/s}^2 \)
- \( t = 45 \, \text{s} \)

3. **Calculate the final speed:**
\[ V = 2000 \cdot \ln\left(\frac{500}{50}\right) - 9.8 \cdot 45 \]
\[ V = 2000 \cdot \ln(10) - 441 \]
\[ V \approx 2000 \cdot 2.302 - 441 \approx 4604 - 441 \approx 4163 \, \text{m/s} \]

4. **Convert to km/s:**
\[ V \approx 4.163 \, \text{km/s} \]

### Summary of Results:
- (i) Minimum rate of fuel consumption to just lift off: **2.45 kg/s**
- (ii) Minimum rate of fuel consumption for an initial acceleration of \( 20 \, \text{m/s}^2 \): **7.45 kg/s**
- Speed of the rocket after all fuel is consumed at \( 10 \, \text{kg/s} \): **4.163 km/s**