The range of a rifle bullet on level ground is 60 m. The range (in m) upon incline of `30^@` is . (Take `g=10m//s^2`)

Explanation

To solve the problem of finding the range of a rifle bullet on an incline of 30 degrees given that the range on level ground is 60 m, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Given Information**:
- The range on level ground (R) = 60 m.
- The angle of incline (θ) = 30 degrees.
- The acceleration due to gravity (g) = 10 m/s².

2. **Use the Range Formula**:
The range of a projectile on level ground is given by the formula:
\[ R = \frac{u^2}{g} \]
where \(u\) is the initial velocity.

3. **Find the Initial Velocity (u)**:
Rearranging the formula to find \(u\):
\[ u^2 = R \cdot g \]
Substituting the values:
\[ u^2 = 60 \cdot 10 = 600 \]
Therefore, \(u = \sqrt{600} = 10\sqrt{6} \, \text{m/s}\).

4. **Calculate the Range on an Incline**:
The range on an incline is given by the formula:
\[ R' = \frac{u^2}{g} \cdot \frac{1}{\cos(\theta)} \cdot (1 + \sin(\theta)) \]
For θ = 30 degrees:
- \(\sin(30^\circ) = \frac{1}{2}\)
- \(\cos(30^\circ) = \frac{\sqrt{3}}{2}\)

Plugging in the values:
\[ R' = \frac{u^2}{g} \cdot \frac{1 + \sin(30^\circ)}{\cos(30^\circ)} \]
\[ R' = \frac{600}{10} \cdot \frac{1 + \frac{1}{2}}{\frac{\sqrt{3}}{2}} \]
\[ R' = 60 \cdot \frac{1.5}{\frac{\sqrt{3}}{2}} = 60 \cdot \frac{3}{\sqrt{3}} = 60\sqrt{3} \]

5. **Final Calculation**:
To find the numerical value:
\[ R' = 60 \cdot \frac{3}{\sqrt{3}} = 60 \cdot \sqrt{3} \approx 60 \cdot 1.732 \approx 103.92 \, \text{m} \]

### Conclusion:
The range of the rifle bullet on an incline of 30 degrees is approximately \( 103.92 \, \text{m} \).