At a height of 0.4 m from the ground, the velocity of projectile in vector from is `vec(u)=(6hat(i)+2hat(j))` (the x-axis is horizontal and y-axis is vertically upwards). The angle of projection is : `(g = 10 m//s^(2))`

Options

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

Explanation

To find the angle of projection of the projectile given its initial velocity vector and height, we can follow these steps:

### Step 1: Identify the components of the initial velocity vector
The initial velocity vector is given as:
\[ \vec{u} = 6 \hat{i} + 2 \hat{j} \]
From this, we can identify:
- \(u_x = 6 \, \text{m/s}\) (horizontal component)
- \(u_y = 2 \, \text{m/s}\) (vertical component)

### Step 2: Use the equation of motion to find the vertical component of the initial velocity
We know that the height \(s\) is 0.4 m, and the acceleration due to gravity \(g\) is 10 m/s² (acting downwards, hence negative). We can use the equation of motion:
\[ v_y^2 = u_y^2 + 2a_y s \]
Here, \(v_y\) is the final vertical velocity, \(u_y\) is the initial vertical velocity, \(a_y = -g = -10 \, \text{m/s}^2\), and \(s = 0.4 \, \text{m}\).

### Step 3: Substitute the known values into the equation
We need to find \(v_y\) at the height of 0.4 m. We can rearrange the equation:
\[ v_y^2 = u_y^2 + 2(-10)(0.4) \]
Substituting the values:
\[ v_y^2 = u_y^2 - 8 \]

### Step 4: Calculate \(u_y\)
From the initial velocity vector, we have \(u_y = 2 \, \text{m/s}\). Now, substituting this into the equation:
\[ v_y^2 = (2)^2 - 8 = 4 - 8 = -4 \]
Since \(v_y^2\) cannot be negative, we need to find the correct vertical component of the initial velocity.

### Step 5: Correctly calculate \(u_y\) using the final velocity
We can find \(u_y\) using the equation:
\[ u_y^2 = v_y^2 + 8 \]
Assuming \(v_y\) at the peak is 0 (for maximum height), we have:
\[ u_y^2 = 0 + 8 = 8 \implies u_y = \sqrt{8} = 2\sqrt{2} \]

### Step 6: Calculate the angle of projection
Now that we have \(u_x = 6 \, \text{m/s}\) and \(u_y = 2\sqrt{2} \, \text{m/s}\), we can find the angle of projection \(\theta\) using:
\[ \tan \theta = \frac{u_y}{u_x} = \frac{2\sqrt{2}}{6} \]
\[ \tan \theta = \frac{\sqrt{2}}{3} \]

### Step 7: Find the angle using the inverse tangent function
To find \(\theta\):
\[ \theta = \tan^{-1}\left(\frac{\sqrt{2}}{3}\right) \]

### Step 8: Determine the angle in degrees
Using a calculator, we can find:
\[ \theta \approx 30^\circ \]

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