At a height `0.4 m` from the ground the velocity of a projectile in vector form is `vec v = (6 hat i+ 2 hat j) ms ^-1`. The angle of projection is

Options

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

Explanation

To find the angle of projection of a projectile given its velocity vector at a height of 0.4 m, 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 and vertical components of the velocity:
- \( v_x = 6 \, \text{m/s} \) (horizontal component)
- \( v_y = 2 \, \text{m/s} \) (vertical component)

### Step 2: Use the kinematic equation to find the initial vertical velocity
We will use the third equation of motion, which relates the final velocity, initial velocity, acceleration, and displacement:
\[ v_y^2 = u_y^2 + 2a s \]
Where:
- \( v_y = 2 \, \text{m/s} \) (final vertical velocity)
- \( u_y \) = initial vertical velocity (which we need to find)
- \( a = -g = -10 \, \text{m/s}^2 \) (acceleration due to gravity, acting downwards)
- \( s = 0.4 \, \text{m} \) (displacement in the vertical direction)

Plugging in the values:
\[ (2)^2 = u_y^2 + 2(-10)(0.4) \]
\[ 4 = u_y^2 - 8 \]
\[ u_y^2 = 4 + 8 = 12 \]
\[ u_y = \sqrt{12} = 2\sqrt{3} \, \text{m/s} \]

### Step 3: Calculate the angle of projection
The angle of projection \( \theta \) can be found using the relationship:
\[ \tan \theta = \frac{u_y}{u_x} \]
Where:
- \( u_x = v_x = 6 \, \text{m/s} \) (since horizontal velocity remains constant)

Substituting the values:
\[ \tan \theta = \frac{2\sqrt{3}}{6} = \frac{\sqrt{3}}{3} \]

### Step 4: Find the angle \( \theta \)
Using the known value of \( \tan \theta \):
\[ \tan \theta = \frac{\sqrt{3}}{3} \implies \theta = 30^\circ \]

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

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