A 5000 kg rocket is set of vertical firing. The exhaust speed is `800" "ms^(-1)`. To give an initial upward acceleration of `20ms^(-2)`, the amount of gas ejected per second to supply the needed thrust will be (take, `g=10ms^(-2)`)

Options

A. 127.5 kg `s^(-1)`
B. 187.5 kg `s^(-1)`
C. 185.5 kg `s^(-1)`
D. 137.5 kg `s^(-1)`

Explanation

To solve the problem of determining the amount of gas ejected per second by the rocket, we will follow these steps:

### Step 1: Identify the known values
- Mass of the rocket, \( m = 5000 \, \text{kg} \)
- Exhaust speed, \( v_r = 800 \, \text{m/s} \)
- Initial upward acceleration, \( a = 20 \, \text{m/s}^2 \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)

### Step 2: Calculate the weight of the rocket
The weight \( W \) of the rocket can be calculated using the formula:
\[ W = m \cdot g \]
Substituting the known values:
\[ W = 5000 \, \text{kg} \cdot 10 \, \text{m/s}^2 = 50000 \, \text{N} \]

### Step 3: Write the equation for thrust force
The thrust force \( F_t \) must overcome both the weight of the rocket and provide the necessary upward acceleration. The net force acting on the rocket can be expressed as:
\[ F_t - W = m \cdot a \]
Rearranging gives:
\[ F_t = W + m \cdot a \]

### Step 4: Substitute the values into the thrust force equation
Substituting the values we have:
\[ F_t = 50000 \, \text{N} + 5000 \, \text{kg} \cdot 20 \, \text{m/s}^2 \]
Calculating the second term:
\[ 5000 \cdot 20 = 100000 \, \text{N} \]
Thus,
\[ F_t = 50000 \, \text{N} + 100000 \, \text{N} = 150000 \, \text{N} \]

### Step 5: Relate thrust force to the mass flow rate of the gas
The thrust force can also be expressed in terms of the mass flow rate \( \frac{dm}{dt} \) and the exhaust speed \( v_r \):
\[ F_t = v_r \cdot \frac{dm}{dt} \]
Setting the two expressions for thrust force equal gives:
\[ 150000 = 800 \cdot \frac{dm}{dt} \]

### Step 6: Solve for the mass flow rate \( \frac{dm}{dt} \)
Rearranging the equation to solve for \( \frac{dm}{dt} \):
\[ \frac{dm}{dt} = \frac{150000}{800} \]
Calculating this gives:
\[ \frac{dm}{dt} = 187.5 \, \text{kg/s} \]

### Final Answer
The amount of gas ejected per second to supply the needed thrust is:
\[ \frac{dm}{dt} = 187.5 \, \text{kg/s} \]

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