A projectile is given an initial velocity of `(hati + sqrt(3) hatj )` m/s, where `hati` is along the ground and `hatj` is along the vertical. Then, the equation of the path of projectile is [ Take g =10 `m//s^(2)]`

Options

A. `y = sqrt(3x ) -5 x^(2)`
B. `y = sqrt(3)x + 5x^(2)`
C. `x = sqrt(3) y + 5x^(2)`
D. `x^(2) =y^(2) +sqrt(3)`

Explanation

To find the equation of the path of a projectile given an initial velocity of \((\hat{i} + \sqrt{3} \hat{j})\) m/s, we can follow these steps:

### Step 1: Identify the components of the initial velocity
The initial velocity vector can be expressed as:
\[ \vec{u} = u_x \hat{i} + u_y \hat{j} \]
where \(u_x = 1 \, \text{m/s}\) and \(u_y = \sqrt{3} \, \text{m/s}\).

### Step 2: Calculate the angle of projection
The angle of projection \(\theta\) can be calculated using the formula:
\[ \tan \theta = \frac{u_y}{u_x} = \frac{\sqrt{3}}{1} = \sqrt{3} \]
Thus, \(\theta = 60^\circ\).

### Step 3: Write the equations of motion
The horizontal and vertical motions can be described by the following equations:
- Horizontal motion (no acceleration):
\[ x = u_x t = 1 \cdot t \]
- Vertical motion (under gravity):
\[ y = u_y t - \frac{1}{2} g t^2 = \sqrt{3} t - \frac{1}{2} \cdot 10 \cdot t^2 \]

### Step 4: Express time \(t\) in terms of \(x\)
From the horizontal motion equation, we can express time \(t\) as:
\[ t = x \]

### Step 5: Substitute \(t\) into the vertical motion equation
Now substitute \(t = x\) into the vertical motion equation:
\[ y = \sqrt{3} x - \frac{1}{2} \cdot 10 \cdot x^2 \]
This simplifies to:
\[ y = \sqrt{3} x - 5 x^2 \]

### Step 6: Final equation of the path
Thus, the equation of the path of the projectile is:
\[ y = \sqrt{3} x - 5 x^2 \]

### Conclusion
The equation of the path of the projectile is \(y = \sqrt{3} x - 5 x^2\).

---