Velocity of a projectile at height `15 m` from ground is `v=(20hati+10hatj)m//s`. Here `hati` is in horizontal direction and `hatj` is vertically upwards. Then
Angle of projectile with ground is

Options

A. `45^(@)`
B. `30^(@)`
C. `37^(@)`
D. `60^(@)`

Answer

To find the angle of the projectile with the ground, we can use the given velocity components at a height of 15 m. The velocity vector is given as:

\[ \mathbf{v} = (20 \hat{i} + 10 \hat{j}) \, \text{m/s} \]

Here, \( \hat{i} \) represents the horizontal direction, and \( \hat{j} \) represents the vertical direction.

### Step 1: Identify the components of the velocity
From the velocity vector, we can identify:
- Horizontal component \( v_x = 20 \, \text{m/s} \)
- Vertical component \( v_y = 10 \, \text{m/s} \)

### Step 2: Use the tangent function to find the angle
The angle \( \theta \) that the projectile makes with the horizontal can be found using the tangent function:

\[ \tan(\theta) = \frac{v_y}{v_x} \]

Substituting the values we have:

\[ \tan(\theta) = \frac{10}{20} = \frac{1}{2} \]

### Step 3: Calculate the angle using the arctangent function
To find \( \theta \), we take the arctangent (inverse tangent) of \( \frac{1}{2} \):

\[ \theta = \tan^{-1}\left(\frac{1}{2}\right) \]

Using a calculator or trigonometric tables, we find:

\[ \theta \approx 26.57^\circ \]

### Step 4: Compare with given options
The options provided are 45 degrees, 30 degrees, 37 degrees, and 60 degrees. The calculated angle \( 26.57^\circ \) does not match any of the options, but we can verify our calculations or check if there was a misunderstanding in the question.

### Final Result
The angle of the projectile with the ground is approximately \( 26.57^\circ \).