A particle is projected in `x-y` plane with `y-`axis along vertical, the point of projection being origin. The equation of projectile is `y = sqrt(3) x - (gx^(2))/(2)`. The angle of projectile is ……………..and initial velocity is ………………… .

Answer

To solve the problem, we need to analyze the given equation of the projectile and compare it with the general equation of projectile motion.

### Step 1: Identify 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 general equation of projectile motion
The general equation of the trajectory of a projectile launched from the origin is:
\[ y = x \tan \theta - \frac{g x^2}{2u^2 \cos^2 \theta} \]

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

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

To find \( \theta \):
\[ \theta = \tan^{-1}(\sqrt{3}) \]
This gives:
\[ \theta = 60^\circ \]

### Step 5: Find the initial velocity
Next, we compare the coefficients of \( x^2 \):
From the equation:
\[ \frac{g}{2u^2 \cos^2 \theta} = 1 \]

Rearranging gives:
\[ 2u^2 \cos^2 \theta = g \]
\[ u^2 \cos^2 \theta = \frac{g}{2} \]

Substituting \( \theta = 60^\circ \):
\[ \cos 60^\circ = \frac{1}{2} \]
Thus:
\[ u^2 \left(\frac{1}{2}\right)^2 = \frac{g}{2} \]
\[ u^2 \cdot \frac{1}{4} = \frac{g}{2} \]

Multiplying both sides by 4:
\[ u^2 = 2g \]

Taking the square root:
\[ u = \sqrt{2g} \]

Assuming \( g \approx 9.8 \, \text{m/s}^2 \):
\[ u = \sqrt{2 \times 9.8} = \sqrt{19.6} \approx 4.43 \, \text{m/s} \]

### Final Answers
- The angle of projection \( \theta \) is \( 60^\circ \).
- The initial velocity \( u \) is approximately \( 4.43 \, \text{m/s} \).