The maximum range of a projectile is 500m. If the particle is thrown up a plane is inclined at an angle of `30^(@)` with the same speed, the distance covered by it along the inclined plane will be:

Options

A. 250 m
B. 500 m
C. 750m
D. 1000 m

Explanation

To solve the problem, we need to find the distance covered by a projectile along an inclined plane when it is thrown with the same speed that gives it a maximum range of 500 m. The incline is at an angle of 30 degrees.

### Step-by-Step Solution:

1. **Understand the Maximum Range**:
The maximum range \( R \) of a projectile is given by the formula:
\[ R = \frac{u^2 \sin(2\theta)}{g} \]
where \( u \) is the initial velocity, \( g \) is the acceleration due to gravity, and \( \theta \) is the angle of projection. For maximum range, \( \theta = 45^\circ \).

2. **Substituting Values**:
Given that the maximum range \( R = 500 \) m, we can substitute this into the equation:
\[ 500 = \frac{u^2 \sin(90^\circ)}{g} \]
Since \( \sin(90^\circ) = 1 \), this simplifies to:
\[ 500 = \frac{u^2}{g} \]

3. **Rearranging for \( u^2 \)**:
Rearranging the equation gives:
\[ u^2 = 500g \]

4. **Finding Distance Along the Inclined Plane**:
When the projectile is thrown up an inclined plane at an angle of \( 30^\circ \), we can use the kinematic equation:
\[ v^2 - u^2 = 2a s \]
Here, \( v = 0 \) (the final velocity at the highest point), \( u^2 = 500g \), and \( a = -g \sin(30^\circ) \).

5. **Substituting Values**:
Since \( \sin(30^\circ) = \frac{1}{2} \), we have:
\[ a = -g \cdot \frac{1}{2} = -\frac{g}{2} \]
Plugging in the values into the kinematic equation:
\[ 0 - (500g) = 2 \left(-\frac{g}{2}\right) s \]
Simplifying this gives:
\[ -500g = -g s \]

6. **Solving for \( s \)**:
Dividing both sides by \( -g \) (assuming \( g \neq 0 \)):
\[ s = 500 \text{ m} \]

### Final Answer:
The distance covered by the projectile along the inclined plane is **500 m**.