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, we need to find the range of a rifle bullet fired at an incline of \(30^\circ\) given that its maximum range on level ground is \(60 \, m\). We will use the equations of projectile motion to derive the solution step by step.

### Step 1: Understand the relationship between range, initial velocity, and angle
The range \(R\) of a projectile launched with an initial velocity \(u\) at an angle \(\theta\) is given by the formula:
\[ R = \frac{u^2 \sin(2\theta)}{g} \]
where \(g\) is the acceleration due to gravity.

### Step 2: Find the initial velocity \(u\)
Given that the maximum range on level ground is \(60 \, m\) at an angle of \(45^\circ\), we can use the range formula:
\[ 60 = \frac{u^2 \sin(90^\circ)}{g} \]
Since \(\sin(90^\circ) = 1\), this simplifies to:
\[ 60 = \frac{u^2}{g} \]
Substituting \(g = 10 \, m/s^2\):
\[ 60 = \frac{u^2}{10} \]
Multiplying both sides by \(10\):
\[ u^2 = 600 \]

### Step 3: Calculate the range at \(30^\circ\)
Now we need to find the range \(R_1\) when the projectile is launched at an angle of \(30^\circ\):
\[ R_1 = \frac{u^2 \sin(2 \times 30^\circ)}{g} \]
Calculating \(\sin(60^\circ)\):
\[ \sin(60^\circ) = \frac{\sqrt{3}}{2} \]
Substituting \(u^2\) and \(g\) into the range formula:
\[ R_1 = \frac{600 \cdot \frac{\sqrt{3}}{2}}{10} \]
This simplifies to:
\[ R_1 = \frac{600 \sqrt{3}}{20} = 30 \sqrt{3} \]

### Step 4: Calculate the numerical value of the range
Now we can calculate \(30 \sqrt{3}\):
Using \(\sqrt{3} \approx 1.732\):
\[ R_1 \approx 30 \times 1.732 = 51.96 \, m \]

### Final Answer
Thus, the range of the rifle bullet upon incline of \(30^\circ\) is approximately \(52 \, m\).