The equation of trajectory of an oblique projectile `y = sqrt(3) x - (g x^(2))/(2)`. The angle of projection is

Options

A. `90^(@)`
B. zero
C. `60^(@)`
D. `30^(@)`

Explanation

To find the angle of projection from the given equation of trajectory of an oblique projectile, we will follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Equation**:
The equation of the trajectory is given as:
\[ y = \sqrt{3} x - \frac{g x^2}{2} \]

2. **Recall the General Form of the Trajectory Equation**:
The standard equation of the trajectory for an oblique projectile is:
\[ y = x \tan \theta - \frac{g x^2}{2 u^2 \cos^2 \theta} \]
Here, \( \theta \) is the angle of projection, \( g \) is the acceleration due to gravity, and \( u \) is the initial velocity.

3. **Compare Coefficients**:
By comparing the coefficients of \( x \) from both equations, we can equate:
\[ \tan \theta = \sqrt{3} \]

4. **Find the Angle of Projection**:
To find \( \theta \), we take the inverse tangent:
\[ \theta = \tan^{-1}(\sqrt{3}) \]

5. **Calculate the Angle**:
We know from trigonometric values that:
\[ \tan 60^\circ = \sqrt{3} \]
Therefore:
\[ \theta = 60^\circ \]

6. **Conclusion**:
The angle of projection is \( 60^\circ \).

### Final Answer:
The angle of projection is \( 60^\circ \).

---