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 step by step, we will analyze the projectile motion of the particle and use the given information to find the angle of projection.

### Step 1: Understand the motion of the particle
The particle is projected from the ground at an angle θ with the horizontal. We know that after 1 second, the particle makes an angle of 45 degrees with the horizontal, and after 2 seconds, its speed is at a minimum (which occurs at the highest point of its trajectory).

### Step 2: Analyze the conditions at t = 1 second
At t = 1 second, the angle of the velocity vector with the horizontal is 45 degrees. This means that the horizontal and vertical components of the velocity are equal at this moment.

Let:
- \( V_x \) = horizontal component of velocity = \( u \cos \theta \)
- \( V_y \) = vertical component of velocity = \( u \sin \theta - g t \)

Since the angle is 45 degrees, we have:
\[ V_y = V_x \]
Thus,
\[ u \sin \theta - g \cdot 1 = u \cos \theta \]
This simplifies to:
\[ u \sin \theta - g = u \cos \theta \]
Rearranging gives:
\[ u \sin \theta - u \cos \theta = g \]
\[ u (\sin \theta - \cos \theta) = g \tag{1} \]

### Step 3: Analyze the conditions at t = 2 seconds
At t = 2 seconds, the particle reaches its highest point, where the vertical component of velocity becomes zero:
\[ V_y = u \sin \theta - g \cdot 2 = 0 \]
This gives us:
\[ u \sin \theta = 2g \tag{2} \]

### Step 4: Relate equations (1) and (2)
From equation (2), we can express \( u \sin \theta \):
\[ u \sin \theta = 20 \quad \text{(using } g = 10 \text{ m/s}^2\text{)} \]

Now, substituting this into equation (1):
\[ 20 - u \cos \theta = 10 \]
This simplifies to:
\[ u \cos \theta = 10 \tag{3} \]

### Step 5: Find the angle of projection
We have two equations now:
1. \( u \sin \theta = 20 \)
2. \( u \cos \theta = 10 \)

Dividing these two equations:
\[ \frac{u \sin \theta}{u \cos \theta} = \frac{20}{10} \]
This simplifies to:
\[ \tan \theta = 2 \]
Thus, the angle of projection θ is:
\[ \theta = \tan^{-1}(2) \]

### Final Answer
The angle of projection of the particle is \( \tan^{-1}(2) \).

---