A particle is projected from gound At a height of 0.4 m from the ground, the velocity of a projective in vector form is `vecv=(6hati+2hatj)m//s` (the x-axis is horizontal and y-axis is vertically upwards). The angle of projection is `(g=10m//s^(2))`

Options

A. `45^(@)`
B. `60^(@)`
C. `30^(@)`
D. `tan^(-1)(3//4)`

Explanation

To find the angle of projection of the particle, we can follow these steps:

### Step 1: Identify the components of the velocity vector
The velocity vector is given as:
\[ \vec{v} = 6 \hat{i} + 2 \hat{j} \, \text{m/s} \]
From this, we can identify the horizontal (x) and vertical (y) components of the velocity:
- \( v_x = 6 \, \text{m/s} \)
- \( v_y = 2 \, \text{m/s} \)

### Step 2: Use the kinematic equation to find the initial vertical velocity
Since the particle is projected from a height of 0.4 m, we can use the kinematic equation:
\[ v_y^2 = u_y^2 - 2gh \]
Rearranging this gives:
\[ u_y^2 = v_y^2 + 2gh \]
Substituting the known values:
- \( v_y = 2 \, \text{m/s} \)
- \( g = 10 \, \text{m/s}^2 \)
- \( h = 0.4 \, \text{m} \)

Calculating \( u_y \):
\[ u_y^2 = (2)^2 + 2 \cdot 10 \cdot 0.4 \]
\[ u_y^2 = 4 + 8 = 12 \]
\[ u_y = \sqrt{12} = 2\sqrt{3} \, \text{m/s} \]

### Step 3: Find the angle of projection
The angle of projection \( \theta \) can be found using the tangent function:
\[ \tan \theta = \frac{u_y}{u_x} \]
Substituting the values:
\[ \tan \theta = \frac{2\sqrt{3}}{6} \]
Simplifying:
\[ \tan \theta = \frac{\sqrt{3}}{3} \]
This implies:
\[ \theta = 30^\circ \]

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

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