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°

Answer

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 projectile motion equations. Here’s a step-by-step solution:

### Step 1: Understand the problem
We know the following:
- The speed of the bullet (initial velocity, \( u \)) = 140 m/s
- The horizontal distance to the target (range, \( R \)) = 1 km = 1000 m
- The acceleration due to gravity (\( g \)) = 9.8 m/s²

### Step 2: Use the range formula for projectile motion
The formula for the range \( R \) of a projectile launched at an angle \( \theta \) with an initial speed \( u \) is given by:

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

### Step 3: Rearrange the formula to find \( \sin(2\theta) \)
We can rearrange the formula to solve for \( \sin(2\theta) \):

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

### Step 4: Substitute the known values
Substituting the known values into the equation:

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

Calculating \( 140^2 \):

\[ 140^2 = 19600 \]

Now substituting this back into the equation:

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

Calculating the right-hand side:

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

### Step 5: Find \( 2\theta \)
Now, we know that:

\[ \sin(2\theta) = 0.5 \]

The angle \( 2\theta \) can be found using the inverse sine function:

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

We know that:

\[ \sin(30^\circ) = 0.5 \]

Thus:

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

### Step 6: Find \( \theta \)
Now, divide by 2 to find \( \theta \):

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

### Conclusion
The angle of projection \( \theta \) required for the bullet to hit the target at a distance of 1 km is:

\[ \theta = 15^\circ \]