A projectile is projected upward with speed `2m//s` on an incline plane of inclination `30^(@)` at an angle of `15^(@)` from the plane. Then the distance along the plane where projectile will fall is :

Options

A. 20 cm
B. 18 cm
C. 26
D. 14 cm

Answer

To solve the problem of finding the distance along the inclined plane where the projectile will fall, we can use the formula for the range of a projectile on an inclined plane. Here's a step-by-step solution:

### Step 1: Identify the parameters
- Initial speed \( u = 2 \, \text{m/s} \)
- Angle of inclination of the plane \( \beta = 30^\circ \)
- Angle of projection from the plane \( \alpha = 15^\circ \)
- Acceleration due to gravity \( g = 10 \, \text{m/s}^2 \)

### Step 2: Use the range formula for projectile motion on an inclined plane
The formula for the range \( R \) along the inclined plane is given by:
\[ R = \frac{2u^2 \sin \alpha \cos(\alpha + \beta)}{g \cos^2 \beta} \]

### Step 3: Calculate \( \sin \alpha \) and \( \cos(\alpha + \beta) \)
- \( \sin 15^\circ \) can be calculated using the sine value:
\[ \sin 15^\circ = \frac{\sqrt{6} - \sqrt{2}}{4} \]
- \( \alpha + \beta = 15^\circ + 30^\circ = 45^\circ \)
- Thus, \( \cos 45^\circ = \frac{1}{\sqrt{2}} \)

### Step 4: Calculate \( \cos^2 \beta \)
- \( \cos 30^\circ = \frac{\sqrt{3}}{2} \)
- Therefore, \( \cos^2 30^\circ = \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4} \)

### Step 5: Substitute the values into the range formula
Now substituting the values into the range formula:
\[ R = \frac{2 \cdot (2^2) \cdot \sin 15^\circ \cdot \cos 45^\circ}{10 \cdot \cos^2 30^\circ} \]
\[ = \frac{2 \cdot 4 \cdot \sin 15^\circ \cdot \frac{1}{\sqrt{2}}}{10 \cdot \frac{3}{4}} \]
\[ = \frac{8 \cdot \sin 15^\circ \cdot \frac{1}{\sqrt{2}}}{10 \cdot \frac{3}{4}} = \frac{8 \cdot \sin 15^\circ \cdot \frac{4}{3}}{10} \]
\[ = \frac{32 \cdot \sin 15^\circ}{30} \]

### Step 6: Calculate \( \sin 15^\circ \) and finalize the calculation
Using \( \sin 15^\circ \approx 0.2588 \):
\[ R = \frac{32 \cdot 0.2588}{30} \approx \frac{8.2816}{30} \approx 0.276 \, \text{m} \]

### Step 7: Convert to centimeters
To convert to centimeters:
\[ R \approx 0.276 \times 100 \approx 27.6 \, \text{cm} \]

### Final Answer
The distance along the plane where the projectile will fall is approximately **27.6 cm**.

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