The parabolic path of a projectile is represented by `y=x/sqrt3-x^(2)/60` in `MKS` units: Its angle of projection is `(g=10ms^(-2))`

Options

A. `30^(@)`
B. `45^(@)`
C. `60^(@)`
D. `90^(@)`

Explanation

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

### Step 1: Understand the given equation
The equation of the projectile's path is given as:
\[ y = \frac{x}{\sqrt{3}} - \frac{x^2}{60} \]

### Step 2: Compare with the standard trajectory equation
The standard form of the trajectory of a projectile is:
\[ y = x \tan \theta - \frac{g x^2}{2 v^2 \cos^2 \theta} \]
where \( g \) is the acceleration due to gravity, \( v \) is the initial velocity, and \( \theta \) is the angle of projection.

### Step 3: Identify the coefficients
From the given equation, we can identify:
- The coefficient of \( x \) is \( \frac{1}{\sqrt{3}} \), which corresponds to \( \tan \theta \).
- The coefficient of \( x^2 \) is \( -\frac{1}{60} \), which corresponds to \( -\frac{g}{2v^2 \cos^2 \theta} \).

### Step 4: Find \( \tan \theta \)
From the comparison, we have:
\[ \tan \theta = \frac{1}{\sqrt{3}} \]

### Step 5: Calculate the angle \( \theta \)
To find \( \theta \), we take the arctangent:
\[ \theta = \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) \]

### Step 6: Determine the angle in degrees
From trigonometric values, we know:
\[ \tan 30^\circ = \frac{1}{\sqrt{3}} \]
Thus,
\[ \theta = 30^\circ \]

### Final Answer
The angle of projection is \( 30^\circ \).

---