A particle is projected with velocity `20sqrt(2)m//s` at `45^(@)` with horizontal. After `1s` , find tangential and normal acceleration of the particle. Also, find radius of curvature of the trajectory at that point. (Take `g=10m//s^(2))`

Explanation

To solve the problem step by step, we will use the following approach:

### Step 1: Break down the initial velocity into components
The initial velocity \( u \) is given as \( 20\sqrt{2} \, \text{m/s} \) at an angle of \( 45^\circ \) with the horizontal.

- The horizontal component \( u_x \) is:
\[ u_x = u \cos(45^\circ) = 20\sqrt{2} \cdot \frac{1}{\sqrt{2}} = 20 \, \text{m/s} \]

- The vertical component \( u_y \) is:
\[ u_y = u \sin(45^\circ) = 20\sqrt{2} \cdot \frac{1}{\sqrt{2}} = 20 \, \text{m/s} \]

### Step 2: Determine the velocity after 1 second
The horizontal velocity remains constant since there is no horizontal acceleration. The vertical velocity changes due to gravity.

- Horizontal velocity \( v_x = u_x = 20 \, \text{m/s} \)

- Vertical velocity after 1 second:
\[ v_y = u_y - g \cdot t = 20 - 10 \cdot 1 = 10 \, \text{m/s} \]

### Step 3: Calculate the total velocity after 1 second
The total velocity \( v \) can be calculated using the Pythagorean theorem:
\[ v = \sqrt{v_x^2 + v_y^2} = \sqrt{(20)^2 + (10)^2} = \sqrt{400 + 100} = \sqrt{500} = 10\sqrt{5} \, \text{m/s} \]

### Step 4: Calculate the tangential acceleration
The tangential acceleration \( a_t \) is due to the change in the vertical component of the velocity. Since the only acceleration acting on the particle is gravity, the tangential acceleration is:
\[ a_t = -g = -10 \, \text{m/s}^2 \]

### Step 5: Calculate the normal acceleration
The normal acceleration \( a_n \) can be calculated using the formula:
\[ a_n = \frac{v^2}{R} \]
where \( R \) is the radius of curvature. We will first find the radius of curvature in the next step.

### Step 6: Calculate the radius of curvature
The radius of curvature \( R \) can be calculated using the formula:
\[ R = \frac{v^2}{g} = \frac{(10\sqrt{5})^2}{10} = \frac{500}{10} = 50 \, \text{m} \]

### Step 7: Calculate the normal acceleration using the radius of curvature
Now we can find the normal acceleration:
\[ a_n = \frac{v^2}{R} = \frac{(10\sqrt{5})^2}{50} = \frac{500}{50} = 10 \, \text{m/s}^2 \]

### Final Results
- Tangential acceleration \( a_t = -10 \, \text{m/s}^2 \)
- Normal acceleration \( a_n = 10 \, \text{m/s}^2 \)
- Radius of curvature \( R = 50 \, \text{m} \)