During the rocket propulsion, gases ejecting with velocity 1 km/s relative to rocket. The rate of fuel consumption is `(m)/(10)` kg/s, where m is the instantaneous mass of the rocket. If air ressitance varies according to equation `f = 0.15` mv, then terminal velocity of the rocket is -

Options

A. 300 m/s
B. 600 m/s
C. 7.92 km/s
D. 11.2 km/s

Answer

To solve the problem of finding the terminal velocity of the rocket, we can follow these steps:

### Step 1: Understand the forces acting on the rocket
The forces acting on the rocket include:
1. The thrust force generated by the ejection of gases.
2. The air resistance (drag) acting against the motion of the rocket.
3. The weight of the rocket (gravitational force).

### Step 2: Write down the thrust force
The thrust force \( F \) generated by the rocket can be expressed as:
\[ F = V_{\text{relative}} \cdot \frac{dm}{dt} \]
where \( V_{\text{relative}} = 1000 \, \text{m/s} \) (the velocity of the gases ejected relative to the rocket) and \( \frac{dm}{dt} = -\frac{m}{10} \) kg/s (the rate of fuel consumption).

### Step 3: Calculate the thrust force
Substituting the values into the thrust force equation:
\[ F = 1000 \cdot \left(-\frac{m}{10}\right) = -100m \, \text{N} \]
(Note: The negative sign indicates that the mass is decreasing.)

### Step 4: Write down the air resistance force
The air resistance (drag) force \( F_d \) is given by:
\[ F_d = 0.15mv \]
where \( v \) is the terminal velocity we want to find.

### Step 5: Write down the weight of the rocket
The weight \( W \) of the rocket is given by:
\[ W = mg \]
where \( g = 10 \, \text{m/s}^2 \).

### Step 6: Set up the equation for terminal velocity
At terminal velocity, the thrust force equals the sum of the air resistance and the weight of the rocket:
\[ -100m = 0.15mv + mg \]
Substituting \( W = mg \):
\[ -100m = 0.15mv + 10m \]

### Step 7: Simplify the equation
Dividing through by \( m \) (assuming \( m \neq 0 \)):
\[ -100 = 0.15v + 10 \]

### Step 8: Solve for terminal velocity \( v \)
Rearranging the equation gives:
\[ -100 - 10 = 0.15v \]
\[ -110 = 0.15v \]
Now, solving for \( v \):
\[ v = \frac{-110}{0.15} = -733.33 \, \text{m/s} \]
(Note: The negative sign indicates the direction of velocity is opposite to the thrust direction, but we are interested in the magnitude.)

### Step 9: Calculate the terminal velocity
Taking the absolute value:
\[ v = 733.33 \, \text{m/s} \]

### Final Answer
The terminal velocity of the rocket is approximately \( 733.33 \, \text{m/s} \).

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