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. `500 km h^(-1)`
B. `1000 km h^(-1)`
C. `1500 km h^(-1)`
D. `2000 km h^(-1)`

Explanation

To solve the problem, we need to find the speed of the burnt gases with respect to an observer on the ground. Let's break down the solution step by step.

### Step 1: Understand the given information
- The speed of the jet airplane (V_plane) = 500 km/h (in one direction).
- The speed of the products of combustion (V_gases) relative to the jet plane = 1500 km/h (in the opposite direction).

### Step 2: Establish the direction of velocities
Since the jet airplane is moving in one direction, we can consider this direction as positive. Therefore:
- The speed of the jet airplane (V_plane) = +500 km/h.
- The speed of the gases relative to the airplane (V_gases relative to plane) = -1500 km/h (because they are ejected in the opposite direction).

### Step 3: Use the formula for relative speed
The formula for relative speed when two objects are moving in opposite directions is:
\[ V_{gases} = V_{plane} + V_{gases \, relative \, to \, plane} \]

### Step 4: Substitute the known values
Substituting the values we have:
\[ V_{gases} = 500 \, \text{km/h} + (-1500 \, \text{km/h}) \]
\[ V_{gases} = 500 \, \text{km/h} - 1500 \, \text{km/h} \]

### Step 5: Calculate the speed of the gases
\[ V_{gases} = -1000 \, \text{km/h} \]

### Step 6: Interpret the result
The negative sign indicates that the speed of the burnt gases is in the opposite direction to that of the jet airplane. Thus, the speed of the burnt gases with respect to the observer on the ground is 1000 km/h in the opposite direction to the airplane's motion.

### Final Answer
The speed of the burnt gases with respect to the observer on the ground is **1000 km/h** in the opposite direction to the jet airplane.

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