A particle is projected from ground with velocity `40sqrt(2)m//s` at `45^(@)`. At time `t=2s`

Options

A. displacement of particle is `100m`
B. vertical component of velocity is `20m//s`
C. velocity makes an angle `tan^(-1)(2)` with vertical
D. particle is at height of `60m` from ground

Answer

To solve the problem, we need to analyze the motion of a particle projected from the ground with a given initial velocity and angle. Here are the steps to find the required quantities at time \( t = 2 \) seconds.

### Step 1: Identify the initial conditions
- Initial velocity \( u = 40\sqrt{2} \, \text{m/s} \)
- Angle of projection \( \theta = 45^\circ \)

### Step 2: Resolve the initial velocity into horizontal and vertical components
- The horizontal component of velocity \( V_x \):
\[ V_x = u \cdot \cos(\theta) = 40\sqrt{2} \cdot \cos(45^\circ) = 40\sqrt{2} \cdot \frac{1}{\sqrt{2}} = 40 \, \text{m/s} \]

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

### Step 3: Calculate the displacements in the x and y directions
- The displacement in the x-direction \( S_x \):
\[ S_x = V_x \cdot t = 40 \cdot 2 = 80 \, \text{m} \]

- The displacement in the y-direction \( S_y \):
\[ S_y = V_y \cdot t + \frac{1}{2} a_y t^2 \]
where \( a_y = -10 \, \text{m/s}^2 \) (acceleration due to gravity).
\[ S_y = 40 \cdot 2 + \frac{1}{2} (-10) \cdot (2^2) = 80 - 20 = 60 \, \text{m} \]

### Step 4: Calculate the resultant displacement
- The resultant displacement \( S \) can be calculated using Pythagoras' theorem:
\[ S = \sqrt{S_x^2 + S_y^2} = \sqrt{80^2 + 60^2} = \sqrt{6400 + 3600} = \sqrt{10000} = 100 \, \text{m} \]

### Step 5: Calculate the vertical component of velocity at \( t = 2 \, \text{s} \)
- The vertical component of velocity \( V_{y} \) at time \( t \):
\[ V_{y} = V_y + a_y \cdot t = 40 - 10 \cdot 2 = 40 - 20 = 20 \, \text{m/s} \]

### Step 6: Calculate the angle of the resultant velocity with respect to the vertical
- The horizontal component remains \( V_x = 40 \, \text{m/s} \) and the vertical component is \( V_y = 20 \, \text{m/s} \).
- The angle \( \phi \) with respect to the vertical can be found using:
\[ \tan(\phi) = \frac{V_x}{V_y} = \frac{40}{20} = 2 \]
Thus, \( \phi = \tan^{-1}(2) \).

### Step 7: Determine the height of the particle
- The height of the particle from the ground is \( S_y = 60 \, \text{m} \).

### Summary of Results
1. Displacement of the particle is \( 100 \, \text{m} \).
2. Vertical component of velocity is \( 20 \, \text{m/s} \).
3. Velocity makes an angle \( \tan^{-1}(2) \) with the vertical.
4. Particle is at a height of \( 60 \, \text{m} \) from the ground.