A particle projected from ground moves at angle `45^(@)` with horizontal one second after projection and speed is minimum two seconds after the projection The angle of projection of particle is [ Neglect the effect of air resistance )

Options

A. `tan^(-1) (3)`
B. `tan^(-1) (2)`
C. `tan^(-1) (sqrt(2))`
D. `tan^(-1) (4)`

Answer

To solve the problem of finding the angle of projection of a particle that is projected from the ground and meets certain conditions, we can follow these steps:

### Step 1: Understand the Problem
We know that:
- The particle is projected at an angle θ with the horizontal.
- After 1 second, the velocity makes an angle of 45 degrees with the horizontal.
- The speed is minimum after 2 seconds.

### Step 2: Analyze the Velocity Components
The velocity of the projectile can be broken down into horizontal and vertical components:
- Horizontal component: \( V_x = U \cos \theta \) (remains constant)
- Vertical component: \( V_y = U \sin \theta - g t \)

### Step 3: Condition After 1 Second
At \( t = 1 \) second, the velocity makes a 45-degree angle with the horizontal. Therefore:
\[ \tan(45^\circ) = \frac{V_y}{V_x} \implies 1 = \frac{V_y}{U \cos \theta} \]
This gives us:
\[ V_y = U \cos \theta \]

Substituting for \( V_y \):
\[ U \sin \theta - g \cdot 1 = U \cos \theta \]
Rearranging gives:
\[ U \sin \theta - U \cos \theta = g \]
Thus:
\[ U (\sin \theta - \cos \theta) = g \quad \text{(Equation 1)} \]

### Step 4: Condition After 2 Seconds
At \( t = 2 \) seconds, the vertical component of the velocity becomes zero (the particle reaches its maximum height):
\[ 0 = U \sin \theta - g \cdot 2 \]
This implies:
\[ U \sin \theta = 2g \quad \text{(Equation 2)} \]

### Step 5: Substitute Equation 2 into Equation 1
From Equation 2, we can express \( \sin \theta \):
\[ \sin \theta = \frac{2g}{U} \]
Substituting this into Equation 1:
\[ U \left(\frac{2g}{U} - \cos \theta\right) = g \]
This simplifies to:
\[ 2g - U \cos \theta = g \]
Rearranging gives:
\[ U \cos \theta = g \quad \text{(Equation 3)} \]

### Step 6: Relate \( \sin \theta \) and \( \cos \theta \)
From Equations 2 and 3:
1. \( \sin \theta = \frac{2g}{U} \)
2. \( \cos \theta = \frac{g}{U} \)

### Step 7: Find \( \tan \theta \)
Now, we can find \( \tan \theta \):
\[ \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\frac{2g}{U}}{\frac{g}{U}} = 2 \]

### Step 8: Calculate the Angle of Projection
Thus, we find:
\[ \theta = \tan^{-1}(2) \]

### Final Answer
The angle of projection of the particle is:
\[ \theta = \tan^{-1}(2) \]