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

Options

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

Explanation

To solve the problem, we need to determine the amount of gas ejected per second (dm/dt) to provide the necessary thrust for the rocket.

### Step-by-Step Solution:

1. **Identify the forces acting on the rocket**:
The forces acting on the rocket are:
- The weight of the rocket, \( W = mg \)
- The thrust produced by the rocket engines, \( T \)

Where:
- \( m = 5000 \, \text{kg} \) (mass of the rocket)
- \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity)

2. **Calculate the weight of the rocket**:
\[ W = mg = 5000 \, \text{kg} \times 10 \, \text{m/s}^2 = 50000 \, \text{N} \]

3. **Determine the required thrust for the rocket**:
The rocket needs to overcome its weight and provide an upward acceleration of \( a = 20 \, \text{m/s}^2 \). The net force required can be calculated using Newton's second law:
\[ F_{\text{net}} = ma \]
The total thrust needed is the weight plus the force required for the upward acceleration:
\[ T = W + F_{\text{net}} = mg + ma = m(g + a) \]
Substituting the values:
\[ T = 5000 \, \text{kg} \times (10 \, \text{m/s}^2 + 20 \, \text{m/s}^2) = 5000 \, \text{kg} \times 30 \, \text{m/s}^2 = 150000 \, \text{N} \]

4. **Relate thrust to the exhaust speed and mass flow rate**:
The thrust produced by the rocket can also be expressed as:
\[ T = V \cdot \frac{dm}{dt} \]
Where:
- \( V = 800 \, \text{m/s} \) (exhaust speed)
- \( \frac{dm}{dt} \) is the mass flow rate (amount of gas ejected per second)

5. **Set the two expressions for thrust equal to each other**:
\[ V \cdot \frac{dm}{dt} = 150000 \, \text{N} \]
Substituting the value of \( V \):
\[ 800 \, \text{m/s} \cdot \frac{dm}{dt} = 150000 \, \text{N} \]

6. **Solve for \( \frac{dm}{dt} \)**:
\[ \frac{dm}{dt} = \frac{150000 \, \text{N}}{800 \, \text{m/s}} = 187.5 \, \text{kg/s} \]

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