Escape velocity of a rocket is 11.2 km/sec. It is released at an angle of `45^(@)` . Its escape velocity is

Options

A. 11.2 m/sec
B. `11.2 sqrt(2)` km/sec
C. 11.2 km/sec
D. 22.3 km/sec

Explanation

To solve the problem regarding the escape velocity of a rocket released at an angle of 45 degrees, we can follow these steps:

### Step 1: Understand Escape Velocity
Escape velocity is the minimum speed needed for an object to break free from the gravitational attraction of a celestial body without any additional propulsion. The formula for escape velocity (V) is given by:

\[ V = \sqrt{\frac{2GM}{R}} \]

where:
- \( G \) is the gravitational constant,
- \( M \) is the mass of the planet,
- \( R \) is the radius of the planet.

### Step 2: Analyze the Given Information
The problem states that the escape velocity of the rocket is 11.2 km/sec. This value is independent of the angle at which the rocket is launched.

### Step 3: Determine the Effect of Launch Angle
The escape velocity is a scalar quantity, which means it does not depend on the direction of the launch. Whether the rocket is launched vertically, horizontally, or at any angle (including 45 degrees), the escape velocity remains the same.

### Step 4: Conclude the Escape Velocity
Since the escape velocity does not depend on the angle of release, the escape velocity of the rocket when released at an angle of 45 degrees is still 11.2 km/sec.

### Final Answer
The escape velocity of the rocket when released at an angle of 45 degrees is:

\[ \text{Escape Velocity} = 11.2 \text{ km/sec} \]