A jet airplane travelling at the speed of ` 500 km ^(-1)` ejects its products of combustion at the speed of ` 1500 km h^(-1)` relative to the jet plane. What is the speed of the burnt gases with respect to observer on the ground ?

Options

A. `– 1000 (km)/h`
B. – 1500 (km)/h
C. `1000 (km)/h`
D. ` 1500 (km)/h`

Explanation

To solve the problem of finding the speed of the burnt gases with respect to an observer on the ground, we can follow these steps:

### Step 1: Identify the velocities
- Let the speed of the jet airplane be \( v_j = 500 \, \text{km/h} \) (in the forward direction).
- The speed of the burnt gases relative to the jet airplane is \( v_{cg/j} = 1500 \, \text{km/h} \) (in the backward direction).

### Step 2: Assign directions
- Since the jet is moving forward, we can consider this direction as positive. Therefore, the backward direction (where the gases are ejected) will be negative.
- Thus, we can express the velocity of the burnt gases relative to the jet as:
\[ v_{cg/j} = -1500 \, \text{km/h} \]

### Step 3: Use the relative velocity formula
- The velocity of the burnt gases with respect to the ground \( v_{cg} \) can be calculated using the formula:
\[ v_{cg} = v_{cg/j} + v_j \]
- Substituting the known values:
\[ v_{cg} = -1500 \, \text{km/h} + 500 \, \text{km/h} \]

### Step 4: Calculate the velocity
- Now, perform the calculation:
\[ v_{cg} = -1500 + 500 = -1000 \, \text{km/h} \]

### Step 5: Interpret the result
- The negative sign indicates that the burnt gases are moving in the opposite direction to the jet airplane. Therefore, the speed of the burnt gases with respect to the observer on the ground is \( 1000 \, \text{km/h} \) in the backward direction.

### Final Answer
The speed of the burnt gases with respect to the observer on the ground is \( 1000 \, \text{km/h} \) in the opposite direction to that of the jet.

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