A bullet fired from a gun with a velocity of `140 ms^(-1)` strikes the ground at the same level as the gun at a distance of 1 km. Find the angle of inclination with the horizontal at which the bullet is fixed. Take `g=9.8ms^(-2)`.

Answer

To find the angle of inclination (θ) at which the bullet is fired from the gun, we can use the formula for the range of a projectile. The range (R) of a projectile launched at an angle θ with an initial velocity (u) is given by:

\[ R = \frac{u^2 \sin(2\theta)}{g} \]

Where:
- R is the range (1 km = 1000 m)
- u is the initial velocity (140 m/s)
- g is the acceleration due to gravity (9.8 m/s²)
- θ is the angle of inclination

### Step 1: Write down the known values
- R = 1000 m
- u = 140 m/s
- g = 9.8 m/s²

### Step 2: Substitute the known values into the range formula
We rearrange the range formula to solve for sin(2θ):

\[ \sin(2\theta) = \frac{R \cdot g}{u^2} \]

Substituting the known values:

\[ \sin(2\theta) = \frac{1000 \cdot 9.8}{(140)^2} \]

### Step 3: Calculate the right-hand side
First, calculate \(140^2\):

\[ 140^2 = 19600 \]

Now substitute this value into the equation:

\[ \sin(2\theta) = \frac{1000 \cdot 9.8}{19600} \]

Calculating the numerator:

\[ 1000 \cdot 9.8 = 9800 \]

Now divide:

\[ \sin(2\theta) = \frac{9800}{19600} = 0.5 \]

### Step 4: Solve for 2θ
Now that we have \(\sin(2\theta) = 0.5\), we can find the angle:

\[ 2\theta = \sin^{-1}(0.5) \]

The value of \(\sin^{-1}(0.5)\) is \(30^\circ\):

\[ 2\theta = 30^\circ \]

### Step 5: Solve for θ
Now divide by 2 to find θ:

\[ \theta = \frac{30^\circ}{2} = 15^\circ \]

### Final Answer
The angle of inclination with the horizontal at which the bullet is fired is:

\[ \theta = 15^\circ \]

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