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 the 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 can use the following kinematic equation to relate the vertical component of the velocity at height \( h \) to the initial vertical velocity \( V_{y0} \):
\[ V_y^2 = V_{y0}^2 - 2g h \]
Where:
- \( V_y = 2 \, \text{m/s} \) (vertical component at height)
- \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity)
- \( h = 0.4 \, \text{m} \) (height)

Rearranging the equation gives:
\[ V_{y0}^2 = V_y^2 + 2gh \]

### Step 3: Substitute the known values
Substituting the known values into the equation:
\[ V_{y0}^2 = (2)^2 + 2 \cdot 10 \cdot 0.4 \]
Calculating this gives:
\[ V_{y0}^2 = 4 + 8 = 12 \]
Thus,
\[ V_{y0} = \sqrt{12} = 2\sqrt{3} \, \text{m/s} \]

### Step 4: Calculate the angle of projection
The angle of projection \( \theta \) can be found using the tangent function:
\[ \tan \theta = \frac{V_{y0}}{V_x} \]
Substituting the values we have:
\[ \tan \theta = \frac{2\sqrt{3}}{6} \]
This simplifies to:
\[ \tan \theta = \frac{\sqrt{3}}{3} \]

### Step 5: Find the angle \( \theta \)
To find \( \theta \), we can use the inverse tangent function:
\[ \theta = \tan^{-1}\left(\frac{\sqrt{3}}{3}\right) \]
This corresponds to:
\[ \theta = 30^\circ \]

### Conclusion
The angle of projection is \( \theta = 30^\circ \).

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