A particle is projected in the x-y plane with y-axis along vertical. Two second after projection the velocity of the particle makers an angle `45^(@)` with the X-axis. Four second after projection. It moves horizontally. Find velocity of projection.

Answer

To solve the problem step by step, we need to analyze the motion of the particle projected in the x-y plane.

### Step 1: Understanding the Problem
The particle is projected at an angle with respect to the horizontal axis. We know two key pieces of information:
1. After 2 seconds, the velocity makes an angle of 45° with the x-axis.
2. After 4 seconds, the particle moves horizontally, meaning the vertical component of its velocity is zero.

### Step 2: Analyzing the Velocity Components
Let the initial velocity of projection be \( U \) at an angle \( \theta \). The components of the initial velocity can be expressed as:
- Horizontal component: \( U_x = U \cos \theta \)
- Vertical component: \( U_y = U \sin \theta \)

### Step 3: Velocity After 2 Seconds
After 2 seconds, the velocity components can be expressed as:
- Horizontal velocity (remains constant): \( V_x = U \cos \theta \)
- Vertical velocity: \( V_y = U \sin \theta - g \cdot t \) where \( g \) is the acceleration due to gravity (approximately \( 10 \, \text{m/s}^2 \)).

At \( t = 2 \) seconds:
\[ V_y = U \sin \theta - 10 \cdot 2 = U \sin \theta - 20 \]

Given that the angle with the x-axis is 45°, we can use the relationship:
\[ \tan(45^\circ) = 1 = \frac{V_y}{V_x} \]
Thus:
\[ V_y = V_x \]
Substituting the expressions for \( V_x \) and \( V_y \):
\[ U \sin \theta - 20 = U \cos \theta \]
Rearranging gives:
\[ U \sin \theta - U \cos \theta = 20 \]
Factoring out \( U \):
\[ U (\sin \theta - \cos \theta) = 20 \quad \text{(1)} \]

### Step 4: Velocity After 4 Seconds
At \( t = 4 \) seconds, the vertical component of the velocity becomes zero:
\[ 0 = U \sin \theta - g \cdot 4 \]
This gives us:
\[ U \sin \theta = 40 \quad \text{(2)} \]

### Step 5: Solving the Equations
Now we have two equations:
1. \( U (\sin \theta - \cos \theta) = 20 \)
2. \( U \sin \theta = 40 \)

From equation (2), we can express \( U \):
\[ U = \frac{40}{\sin \theta} \quad \text{(3)} \]

Substituting (3) into (1):
\[ \frac{40}{\sin \theta} (\sin \theta - \cos \theta) = 20 \]
Simplifying gives:
\[ 40 - \frac{40 \cos \theta}{\sin \theta} = 20 \]
Rearranging:
\[ \frac{40 \cos \theta}{\sin \theta} = 20 \]
Thus:
\[ \frac{\cos \theta}{\sin \theta} = \frac{1}{2} \implies \tan \theta = 2 \]

### Step 6: Finding \( U \)
Using \( \tan \theta = 2 \), we can find \( \sin \theta \) and \( \cos \theta \):
Let \( \sin \theta = 2k \) and \( \cos \theta = k \) for some \( k \):
\[ \sin^2 \theta + \cos^2 \theta = 1 \implies (2k)^2 + k^2 = 1 \implies 4k^2 + k^2 = 1 \implies 5k^2 = 1 \implies k^2 = \frac{1}{5} \implies k = \frac{1}{\sqrt{5}} \]
Thus:
\[ \sin \theta = \frac{2}{\sqrt{5}}, \quad \cos \theta = \frac{1}{\sqrt{5}} \]

Substituting back into equation (3):
\[ U = \frac{40}{\frac{2}{\sqrt{5}}} = 40 \cdot \frac{\sqrt{5}}{2} = 20\sqrt{5} \]

### Final Answer
The velocity of projection \( U \) is \( 20\sqrt{5} \, \text{m/s} \).