The equation of a projectile is `y=sqrt(3)x-(gx^(2))/(2)`
the angle of projection is:-

Options

A. `30^(@)`
B. `60^(@)`
C. `45^(@)`
D. none

Explanation

To find the angle of projection from the given equation of the projectile, we can follow these steps:

### Step 1: Understand the given equation
The equation of the projectile is given as:
\[ y = \sqrt{3}x - \frac{g x^2}{2} \]

### Step 2: Compare with the standard equation of projectile motion
The standard equation of projectile motion is:
\[ y = x \tan(\theta) - \frac{g x^2}{2 v_0^2 \cos^2(\theta)} \]

Here, \( \tan(\theta) \) represents the slope of the trajectory, and \( v_0 \) is the initial velocity.

### Step 3: Identify the components
From the given equation, we can identify:
- The coefficient of \( x \) is \( \sqrt{3} \), which corresponds to \( \tan(\theta) \).
- The term \( -\frac{g x^2}{2} \) corresponds to the term \( -\frac{g x^2}{2 v_0^2 \cos^2(\theta)} \).

### Step 4: Find \( \tan(\theta) \)
From the comparison, we have:
\[ \tan(\theta) = \sqrt{3} \]

### Step 5: Determine the angle \( \theta \)
We know that:
\[ \tan(60^\circ) = \sqrt{3} \]
Thus, we can conclude:
\[ \theta = 60^\circ \]

### Final Answer
The angle of projection is:
\[ \theta = 60^\circ \]

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