An artillary piece which consistently shoots its shells with the same muzzle speed has a maximum range R. To hit a target which is `R//2` from the gun and on the same level, the elevation angle of the gun should be

Options

A. `15^(@)`
B. `45^(@)`
C. `30^(@)`
D. `60^(@)`

Answer

To solve the problem of finding the elevation angle required to hit a target located at a distance of \( R/2 \) from the artillery piece, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Range Formula**:
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.

2. **Maximum Range**:
The maximum range \( R \) occurs when the angle \( \theta \) is \( 45^\circ \). At this angle, the range formula simplifies to:
\[ R = \frac{u^2}{g} \]
From this, we can express the initial velocity \( u \) in terms of \( R \):
\[ u^2 = gR \]

3. **Target Distance**:
We need to find the angle \( \theta \) to hit a target that is at a distance of \( R/2 \). Using the range formula for this distance:
\[ \frac{R}{2} = \frac{u^2 \sin 2\theta}{g} \]

4. **Substituting for \( u^2 \)**:
Substitute \( u^2 = gR \) into the equation:
\[ \frac{R}{2} = \frac{gR \sin 2\theta}{g} \]
This simplifies to:
\[ \frac{R}{2} = R \sin 2\theta \]

5. **Canceling \( R \)**:
Since \( R \) is not zero, we can divide both sides by \( R \):
\[ \frac{1}{2} = \sin 2\theta \]

6. **Finding \( 2\theta \)**:
To find \( 2\theta \), we take the inverse sine:
\[ 2\theta = \sin^{-1}\left(\frac{1}{2}\right) \]
The angle whose sine is \( \frac{1}{2} \) is \( 30^\circ \):
\[ 2\theta = 30^\circ \]

7. **Finding \( \theta \)**:
Now, divide by 2 to find \( \theta \):
\[ \theta = \frac{30^\circ}{2} = 15^\circ \]

### Final Answer:
The elevation angle of the gun should be \( 15^\circ \).