A gun fires a bullet at a speed of `140 ms^(-1)`.If the bullet is to hit a-target at the same level as the gun and at 1km distance, the angle of projection may be

Options

A. `60^(@)" or "30^(@)`
B. `40^(@)" or "50^(@)`
C. `15^(@)" or "75^(@)`
D. `20^(@)" or "70^(@)`

Explanation

To solve the problem of finding the angle of projection for a bullet fired from a gun to hit a target at the same level and at a distance of 1 km, we can use the range formula for projectile motion.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Speed of the bullet, \( u = 140 \, \text{m/s} \)
- Distance to the target, \( R = 1 \, \text{km} = 1000 \, \text{m} \)
- Acceleration due to gravity, \( g = 9.8 \, \text{m/s}^2 \)

2. **Use the Range Formula:**
The range \( R \) of a projectile launched at an angle \( \theta \) is given by the formula:
\[ R = \frac{u^2 \sin(2\theta)}{g} \]
Substituting the known values into the formula:
\[ 1000 = \frac{(140)^2 \sin(2\theta)}{9.8} \]

3. **Rearrange the Equation:**
To find \( \sin(2\theta) \), rearrange the equation:
\[ \sin(2\theta) = \frac{1000 \cdot 9.8}{(140)^2} \]

4. **Calculate the Right Side:**
Calculate \( (140)^2 \):
\[ (140)^2 = 19600 \]
Now substitute this back into the equation:
\[ \sin(2\theta) = \frac{1000 \cdot 9.8}{19600} \]
Simplifying this gives:
\[ \sin(2\theta) = \frac{9800}{19600} = 0.5 \]

5. **Find the Angle \( 2\theta \):**
The sine of an angle is 0.5 at:
\[ 2\theta = 30^\circ \quad \text{or} \quad 2\theta = 150^\circ \]

6. **Calculate \( \theta \):**
From \( 2\theta = 30^\circ \):
\[ \theta = \frac{30^\circ}{2} = 15^\circ \]
From \( 2\theta = 150^\circ \):
\[ \theta = \frac{150^\circ}{2} = 75^\circ \]

7. **Conclusion:**
The possible angles of projection for the bullet to hit the target are:
\[ \theta = 15^\circ \quad \text{or} \quad \theta = 75^\circ \]