The angle of projection of a body is `15^(@)`.. The other angle for which the range is the same as the first one is equal to-

Options

A. `30^(@)`
B. `45^(@)`
C. `60^(@)`
D. `75^(@)`

Answer

To find the other angle of projection that gives the same range as an angle of \(15^\circ\), we can use the property of projectile motion that states the range \(R\) for an angle \(\theta\) is given by the formula:

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

where \(u\) is the initial velocity, \(g\) is the acceleration due to gravity, and \(\theta\) is the angle of projection.

### Step-by-Step Solution:

1. **Identify the given angle**: The angle of projection is given as \( \theta = 15^\circ \).

2. **Calculate \(2\theta\)**:
\[ 2\theta = 2 \times 15^\circ = 30^\circ \]

3. **Write the range formula for \(15^\circ\)**:
\[ R_{15} = \frac{u^2 \sin(30^\circ)}{g} \]

4. **Find the value of \(\sin(30^\circ)\)**:
\[ \sin(30^\circ) = \frac{1}{2} \]
Therefore,
\[ R_{15} = \frac{u^2 \cdot \frac{1}{2}}{g} = \frac{u^2}{2g} \]

5. **Determine the other angle**: We need to find another angle \(\theta'\) such that:
\[ R_{\theta'} = R_{15} \]
This means:
\[ R_{\theta'} = \frac{u^2 \sin(2\theta')}{g} \]
Setting the two ranges equal gives:
\[ \frac{u^2 \sin(2\theta')}{g} = \frac{u^2}{2g} \]

6. **Cancel \(u^2\) and \(g\)** (assuming \(u \neq 0\) and \(g \neq 0\)):
\[ \sin(2\theta') = \frac{1}{2} \]

7. **Find the angles for which \(\sin(2\theta') = \frac{1}{2}\)**:
The general solutions for \(\sin x = \frac{1}{2}\) are:
\[ x = 30^\circ + n \cdot 360^\circ \quad \text{and} \quad x = 150^\circ + n \cdot 360^\circ \]
where \(n\) is any integer.

8. **Substituting back for \(2\theta'\)**:
- For \(2\theta' = 30^\circ\):
\[ \theta' = \frac{30^\circ}{2} = 15^\circ \]
- For \(2\theta' = 150^\circ\):
\[ \theta' = \frac{150^\circ}{2} = 75^\circ \]

9. **Conclusion**: The other angle of projection that gives the same range as \(15^\circ\) is:
\[ \theta' = 75^\circ \]

### Final Answer:
The other angle for which the range is the same as the first one is \(75^\circ\).