A bullet is fired from horizontal ground at some angle passes through the point (3R/4 ,R/4), where `R` is the range of the bullet. Assume point of the fire to be origin and the bullet moves in x-y plane with x-axis horizontal and y-axis vertically upwards. Then angle of projection is

Options

A. `30^@`
B. `37^@`
C. `53^@`
D. none

Explanation

To solve the problem, we need to find the angle of projection of a bullet fired from the horizontal ground that passes through the point \((\frac{3R}{4}, \frac{R}{4})\), where \(R\) is the range of the bullet.

### Step-by-Step Solution:

1. **Understanding the Projectile Motion**:
The bullet follows a parabolic trajectory. The horizontal and vertical motions can be described using the equations of projectile motion. The horizontal range \(R\) is given by:
\[ R = \frac{u^2 \sin 2\theta}{g} \]
where \(u\) is the initial velocity, \(\theta\) is the angle of projection, and \(g\) is the acceleration due to gravity.

2. **Equation of Trajectory**:
The equation of the trajectory of the projectile is given by:
\[ y = x \tan \theta - \frac{g x^2}{2u^2 \cos^2 \theta} \]
We will use this equation to find the angle \(\theta\).

3. **Substituting the Given Point**:
We know the bullet passes through the point \((\frac{3R}{4}, \frac{R}{4})\). We will substitute \(x = \frac{3R}{4}\) and \(y = \frac{R}{4}\) into the trajectory equation:
\[ \frac{R}{4} = \frac{3R}{4} \tan \theta - \frac{g \left(\frac{3R}{4}\right)^2}{2u^2 \cos^2 \theta} \]

4. **Simplifying the Equation**:
Rearranging the equation gives:
\[ \frac{R}{4} + \frac{g \left(\frac{3R}{4}\right)^2}{2u^2 \cos^2 \theta} = \frac{3R}{4} \tan \theta \]
Dividing through by \(R\) (assuming \(R \neq 0\)):
\[ \frac{1}{4} + \frac{g \left(\frac{3}{4}\right)^2}{2u^2 \cos^2 \theta} = \frac{3}{4} \tan \theta \]

5. **Using the Range Formula**:
From the range formula, we can express \(u^2\) in terms of \(R\) and \(\theta\):
\[ u^2 = \frac{gR}{\sin 2\theta} \]
Substituting this into our equation gives:
\[ \frac{1}{4} + \frac{g \left(\frac{3}{4}\right)^2}{2 \left(\frac{gR}{\sin 2\theta}\right) \cos^2 \theta} = \frac{3}{4} \tan \theta \]

6. **Further Simplifying**:
Simplifying the left-hand side:
\[ \frac{1}{4} + \frac{\frac{9g}{16}}{2 \frac{gR}{\sin 2\theta} \cos^2 \theta} = \frac{3}{4} \tan \theta \]
This simplifies to:
\[ \frac{1}{4} + \frac{9 \sin 2\theta}{32R \cos^2 \theta} = \frac{3}{4} \tan \theta \]

7. **Finding \(\tan \theta\)**:
Rearranging gives:
\[ \frac{9 \sin 2\theta}{32R \cos^2 \theta} = \frac{3}{4} \tan \theta - \frac{1}{4} \]
Solving this equation will yield \(\tan \theta\).

8. **Final Calculation**:
After solving, we find:
\[ \tan \theta = \frac{4}{3} \]
Thus, the angle of projection \(\theta\) can be calculated as:
\[ \theta = \tan^{-1}\left(\frac{4}{3}\right) \]
This gives approximately:
\[ \theta \approx 53^\circ \]

### Final Answer:
The angle of projection is \(53^\circ\).