(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, we will break it down into two parts as specified in the question.

### Part (a)

#### (i) Minimum Rate of Fuel Consumption to Just Lift Off

1. **Identify the forces acting on the rocket:**
- The thrust force (F_t) generated by the rocket must equal the weight (W) of the rocket to just lift off.
- Weight (W) = mg, where m is the total mass of the rocket (rocket mass + fuel mass) and g is the acceleration due to gravity.

2. **Calculate the total mass of the rocket:**
- Mass of the rocket = 50 kg
- Mass of the fuel = 450 kg
- Total mass (m) = 50 kg + 450 kg = 500 kg

3. **Write the equation for thrust:**
- Thrust force (F_t) = exhaust velocity (V_e) × rate of fuel consumption (-dm/dt)
- To just lift off: F_t = W
- Therefore, V_e × (-dm/dt) = mg

4. **Rearranging the equation to find the minimum rate of fuel consumption:**
\[ -\frac{dm}{dt} = \frac{mg}{V_e} \]

5. **Substituting the values:**
- g = 9.8 m/s²
- V_e = 2000 m/s (2 km/s)
\[ -\frac{dm}{dt} = \frac{500 \times 9.8}{2000} = \frac{4900}{2000} = 2.45 \text{ kg/s} \]

Thus, the minimum rate of fuel consumption to just lift off is **2.45 kg/s**.

#### (ii) Minimum Rate of Fuel Consumption for Initial Acceleration of 20 m/s²

1. **Identify the net force for the desired acceleration:**
- According to Newton's second law: F_net = ma
- The thrust must overcome both the weight and provide the necessary acceleration.

2. **Write the equation for thrust with acceleration:**
\[ F_t - W = ma \]
- Therefore, F_t = W + ma

3. **Substituting the values:**
- W = mg = 500 kg × 9.8 m/s² = 4900 N
- a = 20 m/s²
\[ F_t = 4900 + 500 \times 20 = 4900 + 10000 = 14900 \text{ N} \]

4. **Using the thrust equation:**
\[ F_t = V_e \times (-dm/dt) \]
\[ 14900 = 2000 \times (-dm/dt) \]

5. **Rearranging to find the rate of fuel consumption:**
\[ -\frac{dm}{dt} = \frac{14900}{2000} = 7.45 \text{ kg/s} \]

Thus, the minimum rate of fuel consumption for an initial acceleration of 20 m/s² is **7.45 kg/s**.

### Part (b)

1. **Determine the final speed of the rocket after all fuel is consumed:**
- The initial mass (m₀) = 500 kg
- The mass of the rocket after fuel consumption (m) = 50 kg (only the rocket remains)
- Rate of fuel consumption = 10 kg/s
- Total fuel = 450 kg
- Time to consume all fuel = 450 kg / 10 kg/s = 45 s

2. **Using the rocket equation:**
\[ v = V_e \ln\left(\frac{m_0}{m}\right) - gt \]
- Where:
- V_e = 2000 m/s
- m₀ = 500 kg
- m = 50 kg
- g = 9.8 m/s²
- t = 45 s

3. **Substituting the values:**
\[ v = 2000 \ln\left(\frac{500}{50}\right) - 9.8 \times 45 \]
\[ v = 2000 \ln(10) - 441 \]
- Using ln(10) ≈ 2.303:
\[ v = 2000 \times 2.303 - 441 \]
\[ v = 4606 - 441 = 4165 \text{ m/s} \]

Thus, the speed of the rocket when the rate of consumption of fuel is 10 kg/s after all the fuel is consumed is approximately **4165 m/s** or **4.165 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 initial acceleration of 20 m/s²: **7.45 kg/s**
- (b) Speed of the rocket after all fuel is consumed: **4.165 km/s**