A particle is projected from ground with velocity `40(sqrt2) m//s` at `45^@`. Find
(a) velocity and
(b) displacement of the particle after 2 s. `(g = 10 m//s^2)`

Explanation

To solve the problem, we will break it down into steps to find both the velocity and displacement of the particle after 2 seconds.

### Step 1: Determine the initial velocity components
The initial velocity \( u \) is given as \( 40\sqrt{2} \, \text{m/s} \) at an angle of \( 45^\circ \).

- The horizontal component of the initial velocity \( u_x \):
\[ u_x = u \cdot \cos(45^\circ) = 40\sqrt{2} \cdot \frac{1}{\sqrt{2}} = 40 \, \text{m/s} \]

- The vertical component of the initial velocity \( u_y \):
\[ u_y = u \cdot \sin(45^\circ) = 40\sqrt{2} \cdot \frac{1}{\sqrt{2}} = 40 \, \text{m/s} \]

### Step 2: Calculate the velocity after 2 seconds
- The horizontal velocity \( v_x \) remains constant since there is no acceleration in the horizontal direction:
\[ v_x = u_x = 40 \, \text{m/s} \]

- The vertical velocity \( v_y \) after 2 seconds can be calculated using the first equation of motion:
\[ v_y = u_y + a_y \cdot t \]
where \( a_y = -g = -10 \, \text{m/s}^2 \) (acceleration due to gravity) and \( t = 2 \, \text{s} \):
\[ v_y = 40 + (-10) \cdot 2 = 40 - 20 = 20 \, \text{m/s} \]

### Step 3: Write the velocity vector
The velocity vector \( \vec{v} \) after 2 seconds is:
\[ \vec{v} = v_x \hat{i} + v_y \hat{j} = 40 \hat{i} + 20 \hat{j} \, \text{m/s} \]

### Step 4: Calculate the displacement after 2 seconds
- The horizontal displacement \( x \) can be calculated as:
\[ x = u_x \cdot t = 40 \cdot 2 = 80 \, \text{m} \]

- The vertical displacement \( y \) can be calculated using the second equation of motion:
\[ y = u_y \cdot t + \frac{1}{2} a_y t^2 \]
\[ y = 40 \cdot 2 + \frac{1}{2} \cdot (-10) \cdot (2^2) = 80 - 20 = 60 \, \text{m} \]

### Step 5: Write the displacement vector
The displacement vector \( \vec{s} \) after 2 seconds is:
\[ \vec{s} = x \hat{i} + y \hat{j} = 80 \hat{i} + 60 \hat{j} \, \text{m} \]

### Step 6: Calculate the magnitude of the displacement
The magnitude of the displacement \( S \) can be calculated using the Pythagorean theorem:
\[ S = \sqrt{x^2 + y^2} = \sqrt{80^2 + 60^2} = \sqrt{6400 + 3600} = \sqrt{10000} = 100 \, \text{m} \]

### Final Answers
- (a) The velocity after 2 seconds is \( \vec{v} = 40 \hat{i} + 20 \hat{j} \, \text{m/s} \)
- (b) The displacement after 2 seconds is \( \vec{s} = 80 \hat{i} + 60 \hat{j} \, \text{m} \) with a magnitude of \( 100 \, \text{m} \)