What will be the equation for the path of a projectile if it is fired with initial velocity of `(2hati+3hatj)`. Here `hatu and hatj` are unit vectors along X-axis and Y-axis `(g=10ms^(-2))`.

Answer

To find the equation for the path of a projectile fired with an initial velocity of \( \mathbf{u} = 2\hat{i} + 3\hat{j} \), we will follow these steps:

### Step 1: Identify the components of the initial velocity
The initial velocity vector can be broken down into its components:
- \( u_x = 2 \, \text{m/s} \) (horizontal component)
- \( u_y = 3 \, \text{m/s} \) (vertical component)

### Step 2: Calculate the angle of projection
Using the components of the initial velocity, we can find the angle \( \theta \) of projection using the tangent function:
\[ \tan \theta = \frac{u_y}{u_x} = \frac{3}{2} \]

### Step 3: Calculate the magnitude of the initial velocity
The magnitude of the initial velocity \( u \) can be calculated using the Pythagorean theorem:
\[ u = \sqrt{u_x^2 + u_y^2} = \sqrt{2^2 + 3^2} = \sqrt{4 + 9} = \sqrt{13} \]

### Step 4: Write the equation of the trajectory
The general equation of the trajectory of a projectile is given by:
\[ y = x \tan \theta - \frac{g x^2}{2 u^2 \cos^2 \theta} \]
Substituting \( g = 10 \, \text{m/s}^2 \) and \( u = \sqrt{13} \):
1. Calculate \( \tan \theta \):
\[ \tan \theta = \frac{3}{2} \]
2. Calculate \( \cos^2 \theta \):
\[ \cos^2 \theta = \frac{u_x^2}{u^2} = \frac{2^2}{13} = \frac{4}{13} \]

Now substituting these values into the trajectory equation:
\[ y = x \left(\frac{3}{2}\right) - \frac{10 x^2}{2 \cdot 13 \cdot \frac{4}{13}} \]
This simplifies to:
\[ y = \frac{3}{2} x - \frac{10 x^2}{8} = \frac{3}{2} x - \frac{5}{4} x^2 \]

### Step 5: Rearranging the equation
To express the equation in a standard form, we can multiply through by 4 to eliminate the fraction:
\[ 4y = 6x - 5x^2 \]
Rearranging gives us:
\[ 5x^2 - 6x + 4y = 0 \]

### Final Equation
Thus, the equation for the path of the projectile is:
\[ 4y = 6x - 5x^2 \]

---