A particle is projected from ground with velocity `50 m//s` at `37^@` from horizontal. Find velocity and displacement after 2 s. `sin 37^@ = 3/5`.

Explanation

To solve the problem of finding the velocity and displacement of a particle projected from the ground with a given initial velocity and angle, we can follow these steps:

### Step 1: Determine the initial velocity components
Given:
- Initial velocity, \( u = 50 \, \text{m/s} \)
- Angle of projection, \( \theta = 37^\circ \)
- \( \sin 37^\circ = \frac{3}{5} \)

We can calculate the horizontal (x) and vertical (y) components of the initial velocity using trigonometric functions:

\[ u_x = u \cdot \cos \theta \]
\[ u_y = u \cdot \sin \theta \]

Using the value of \( \sin 37^\circ \) and the relationship \( \cos^2 \theta + \sin^2 \theta = 1 \), we can find \( \cos 37^\circ \):

\[ \cos 37^\circ = \sqrt{1 - \left(\frac{3}{5}\right)^2} = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5} \]

Now, substituting the values:

\[ u_x = 50 \cdot \frac{4}{5} = 40 \, \text{m/s} \]
\[ u_y = 50 \cdot \frac{3}{5} = 30 \, \text{m/s} \]

### Step 2: Write the initial velocity vector
The initial velocity vector can be expressed as:

\[ \mathbf{u} = 40 \hat{i} + 30 \hat{j} \, \text{m/s} \]

### Step 3: Determine the acceleration components
Since the only acceleration acting on the particle is due to gravity, we have:

\[ \mathbf{a} = 0 \hat{i} - 10 \hat{j} \, \text{m/s}^2 \]

### Step 4: Calculate the velocity after 2 seconds
Using the equation of motion:

\[ \mathbf{v} = \mathbf{u} + \mathbf{a} t \]

Substituting the known values:

\[ \mathbf{v} = (40 \hat{i} + 30 \hat{j}) + (0 \hat{i} - 10 \hat{j}) \cdot 2 \]
\[ \mathbf{v} = 40 \hat{i} + 30 \hat{j} - 20 \hat{j} \]
\[ \mathbf{v} = 40 \hat{i} + 10 \hat{j} \, \text{m/s} \]

### Step 5: Calculate the displacement after 2 seconds
Using the equation of motion for displacement:

\[ \mathbf{s} = \mathbf{u} t + \frac{1}{2} \mathbf{a} t^2 \]

Substituting the known values:

\[ \mathbf{s} = (40 \hat{i} + 30 \hat{j}) \cdot 2 + \frac{1}{2} (0 \hat{i} - 10 \hat{j}) \cdot (2^2) \]
\[ \mathbf{s} = (80 \hat{i} + 60 \hat{j}) + \frac{1}{2} (0 \hat{i} - 40 \hat{j}) \]
\[ \mathbf{s} = 80 \hat{i} + 60 \hat{j} - 20 \hat{j} \]
\[ \mathbf{s} = 80 \hat{i} + 40 \hat{j} \, \text{m} \]

### Final Results
- **Velocity after 2 seconds:** \( \mathbf{v} = 40 \hat{i} + 10 \hat{j} \, \text{m/s} \)
- **Displacement after 2 seconds:** \( \mathbf{s} = 80 \hat{i} + 40 \hat{j} \, \text{m} \)