A particle is projected from ground at angle `45^@` with initial velocity `20(sqrt2) m//s`. Find
(a) change in velocity,
(b) magnitude of average velocity in a time interval from `t=0` to `t=3s`.

Explanation

To solve the problem step by step, let's break it down into two parts: (a) finding the change in velocity and (b) finding the magnitude of average velocity over the time interval from \( t = 0 \) to \( t = 3 \) seconds.

### Step-by-Step Solution

**Part (a): Change in Velocity**

1. **Determine the initial velocity components:**
The initial velocity \( u \) is given as \( 20\sqrt{2} \, \text{m/s} \) at an angle of \( 45^\circ \).
- The horizontal component \( u_x = u \cos(45^\circ) = 20\sqrt{2} \cdot \frac{1}{\sqrt{2}} = 20 \, \text{m/s} \).
- The vertical component \( u_y = u \sin(45^\circ) = 20\sqrt{2} \cdot \frac{1}{\sqrt{2}} = 20 \, \text{m/s} \).

Thus, the initial velocity vector is:
\[ \mathbf{u} = 20 \hat{i} + 20 \hat{j} \, \text{m/s} \]

2. **Calculate the final velocity after 3 seconds:**
The acceleration due to gravity \( a \) is \( -10 \, \text{m/s}^2 \) (acting downwards).
- The final velocity in the horizontal direction remains unchanged: \( v_x = u_x = 20 \, \text{m/s} \).
- The final velocity in the vertical direction is given by:
\[ v_y = u_y + a t = 20 + (-10)(3) = 20 - 30 = -10 \, \text{m/s} \]

Thus, the final velocity vector is:
\[ \mathbf{v} = 20 \hat{i} - 10 \hat{j} \, \text{m/s} \]

3. **Calculate the change in velocity:**
The change in velocity \( \Delta \mathbf{v} \) is given by:
\[ \Delta \mathbf{v} = \mathbf{v} - \mathbf{u} = (20 \hat{i} - 10 \hat{j}) - (20 \hat{i} + 20 \hat{j}) = 0 \hat{i} - 30 \hat{j} = -30 \hat{j} \, \text{m/s} \]

**Part (b): Magnitude of Average Velocity**

1. **Calculate the displacement during the time interval:**
The displacement \( \mathbf{s} \) can be calculated using the formula:
\[ \mathbf{s} = \mathbf{u} t + \frac{1}{2} \mathbf{a} t^2 \]
where \( \mathbf{a} = 0 \hat{i} - 10 \hat{j} \, \text{m/s}^2 \).
- The displacement in the horizontal direction:
\[ s_x = u_x t = 20 \cdot 3 = 60 \, \text{m} \]
- The displacement in the vertical direction:
\[ s_y = u_y t + \frac{1}{2} a_y t^2 = 20 \cdot 3 + \frac{1}{2} (-10)(3^2) = 60 - 45 = 15 \, \text{m} \]

Thus, the displacement vector is:
\[ \mathbf{s} = 60 \hat{i} + 15 \hat{j} \, \text{m} \]

2. **Calculate the average velocity:**
The average velocity \( \mathbf{v}_{\text{avg}} \) is given by:
\[ \mathbf{v}_{\text{avg}} = \frac{\Delta \mathbf{s}}{t} = \frac{60 \hat{i} + 15 \hat{j}}{3} = 20 \hat{i} + 5 \hat{j} \, \text{m/s} \]

3. **Find the magnitude of the average velocity:**
The magnitude of \( \mathbf{v}_{\text{avg}} \) is calculated as:
\[ |\mathbf{v}_{\text{avg}}| = \sqrt{(20)^2 + (5)^2} = \sqrt{400 + 25} = \sqrt{425} \approx 20.62 \, \text{m/s} \]

### Final Answers
(a) Change in velocity: \( -30 \hat{j} \, \text{m/s} \)
(b) Magnitude of average velocity: \( \approx 20.62 \, \text{m/s} \)