A fighter plane is moving in a vertical circle of radius 'r'. Its minimum velocity at the highest point of the circle will be

Options

A. `sqrt(3gr)`
B. `sqrt(2 gr)`
C. `sqrt(gr)`
D. `sqrt(gr//2)`

Answer

To find the minimum velocity of a fighter plane moving in a vertical circle at the highest point, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Forces at the Highest Point**:
At the highest point of the vertical circle, the forces acting on the fighter plane are its weight (mg) acting downwards and the centripetal force required to keep it moving in a circle.

2. **Centripetal Force Requirement**:
The centripetal force (F_c) needed to keep the plane in circular motion is provided by the gravitational force at the highest point. Thus, we can write:
\[ F_c = \frac{mv^2}{r} \]
where \( m \) is the mass of the plane, \( v \) is its velocity, and \( r \) is the radius of the circle.

3. **Setting Up the Equation**:
At the highest point, the gravitational force must be equal to or greater than the centripetal force required to keep the plane in circular motion. Therefore, we can set up the inequality:
\[ mg \geq \frac{mv^2}{r} \]

4. **Canceling Mass**:
Since the mass \( m \) appears on both sides of the equation, we can cancel it out (assuming \( m \neq 0 \)):
\[ g \geq \frac{v^2}{r} \]

5. **Rearranging the Equation**:
Rearranging the equation gives us:
\[ v^2 \leq gr \]

6. **Finding Minimum Velocity**:
To find the minimum velocity at the highest point, we take the equality case:
\[ v^2 = gr \]
Taking the square root of both sides, we find:
\[ v = \sqrt{gr} \]

7. **Conclusion**:
Therefore, the minimum velocity of the fighter plane at the highest point of the vertical circle is:
\[ v = \sqrt{gr} \]

### Final Answer:
The minimum velocity at the highest point of the circle is \( v = \sqrt{gr} \).