A particle is fired horizontally form an inclined plane of inclination `30^@` with horizontal with speed `50 ms^-1`. If `g=10 ms^-2` , the range measured along the incline is

Options

A. `500 m`
B. `1000/3 m`
C. `200 (sqrt2)` m
D. `100 (sqrt3)` m

Explanation

To solve the problem of finding the range of a particle fired horizontally from an inclined plane, we can follow these steps:

### Step 1: Identify the given values
- Initial speed of the particle, \( u = 50 \, \text{m/s} \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)
- Angle of inclination of the plane, \( \beta = 30^\circ \)
- The angle of projection, \( \alpha = 0^\circ \) (since the particle is fired horizontally).

### Step 2: Use the formula for range along an incline
The formula for the range \( R \) of a projectile launched from an inclined plane is given by:

\[ R = \frac{u^2}{g \cos^2 \beta} \left( \sin(2\alpha + \beta) + \sin \beta \right) \]

### Step 3: Substitute the values into the formula
Substituting the known values into the formula:

\[ R = \frac{50^2}{10 \cos^2(30^\circ)} \left( \sin(2 \cdot 0^\circ + 30^\circ) + \sin(30^\circ) \right) \]

### Step 4: Calculate each component
1. Calculate \( u^2 \):
\[ u^2 = 50^2 = 2500 \]

2. Calculate \( g \cos^2(30^\circ) \):
- \( \cos(30^\circ) = \frac{\sqrt{3}}{2} \)
- Therefore, \( \cos^2(30^\circ) = \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4} \)
- Thus, \( g \cos^2(30^\circ) = 10 \cdot \frac{3}{4} = 7.5 \)

3. Calculate \( \sin(30^\circ) \):
\[ \sin(30^\circ) = \frac{1}{2} \]

4. Calculate \( \sin(2 \cdot 0^\circ + 30^\circ) \):
\[ \sin(30^\circ) = \frac{1}{2} \]

5. Now substitute these values back into the range formula:
\[ R = \frac{2500}{7.5} \left( \frac{1}{2} + \frac{1}{2} \right) \]

### Step 5: Simplify the expression
\[ R = \frac{2500}{7.5} \cdot 1 = \frac{2500}{7.5} = \frac{25000}{75} = \frac{1000}{3} \, \text{m} \]

### Final Answer
The range measured along the incline is:
\[ R = \frac{1000}{3} \, \text{m} \approx 333.33 \, \text{m} \]

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