A particle is projected with a velocity of 50 m/s at `37^@` with horizontal. Find velocity, displacement and co-ordinates of the particle (w.r.t. the starting point) after 2 s.
Given, `g=10m//s^2, sin 37^@ = 0.6 and cos 37^@ = 0.8`.

Explanation

To solve the problem step by step, we will calculate the horizontal and vertical components of the initial velocity, the velocity after 2 seconds, the displacement after 2 seconds, and finally the coordinates of the particle with respect to the starting point.

### Step 1: Calculate the horizontal and vertical components of the initial velocity

Given:
- Initial velocity \( u = 50 \, \text{m/s} \)
- Angle \( \theta = 37^\circ \)

The horizontal component of the velocity \( u_x \) is given by:
\[ u_x = u \cdot \cos(\theta) = 50 \cdot \cos(37^\circ) = 50 \cdot 0.8 = 40 \, \text{m/s} \]

The vertical component of the velocity \( u_y \) is given by:
\[ u_y = u \cdot \sin(\theta) = 50 \cdot \sin(37^\circ) = 50 \cdot 0.6 = 30 \, \text{m/s} \]

### Step 2: Determine the acceleration due to gravity

The acceleration due to gravity \( g \) is given as:
\[ g = 10 \, \text{m/s}^2 \]
This acts downwards, so we can represent it as:
\[ \text{Acceleration} = -g \, \hat{j} = -10 \, \hat{j} \, \text{m/s}^2 \]

### Step 3: Calculate the velocity after 2 seconds

Using the equation of motion:
\[ v = u + at \]
where \( t = 2 \, \text{s} \).

The horizontal component of the velocity after 2 seconds:
\[ v_x = u_x + 0 = 40 \, \text{m/s} \quad (\text{no horizontal acceleration}) \]

The vertical component of the velocity after 2 seconds:
\[ v_y = u_y - g \cdot t = 30 - 10 \cdot 2 = 30 - 20 = 10 \, \text{m/s} \]

Thus, the velocity vector after 2 seconds is:
\[ \vec{v} = 40 \, \hat{i} + 10 \, \hat{j} \, \text{m/s} \]

### Step 4: Calculate the displacement after 2 seconds

Using the equation for displacement:
\[ \vec{s} = \vec{u} t + \frac{1}{2} \vec{a} t^2 \]

Calculating the displacement:
\[ \vec{s} = (40 \, \hat{i} + 30 \, \hat{j}) \cdot 2 + \frac{1}{2} (-10 \, \hat{j}) \cdot (2^2) \]
Calculating each term:
\[ \vec{s} = (80 \, \hat{i} + 60 \, \hat{j}) + \frac{1}{2} (-10) \cdot 4 \, \hat{j} \]
\[ \vec{s} = 80 \, \hat{i} + 60 \, \hat{j} - 20 \, \hat{j} \]
\[ \vec{s} = 80 \, \hat{i} + 40 \, \hat{j} \, \text{m} \]

### Step 5: Determine the coordinates of the particle

The coordinates of the particle with respect to the starting point after 2 seconds are:
\[ (x, y) = (80, 40) \]

### Final Results

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