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 will analyze the motion of the particle in the x-y plane, taking into account the information given about its velocity at specific times.

### Step 1: Understanding the Motion
The particle is projected in the x-y plane, where the y-axis is vertical. We need to find the initial velocity of the particle given that:
- After 2 seconds, the velocity makes a 45-degree angle with the x-axis.
- After 4 seconds, the particle moves horizontally.

### Step 2: Analyze the Velocity Components
At 2 seconds, the velocity \( V \) makes an angle of 45 degrees with the x-axis. This means that the horizontal component \( V_x \) and the vertical component \( V_y \) of the velocity are equal:
\[ V_x = V_y \]

### Step 3: Determine the Horizontal and Vertical Components
Since there is no horizontal acceleration (the only acceleration is due to gravity acting downward), the horizontal component of the initial velocity \( u_x \) remains constant:
\[ u_x = V_x \]

### Step 4: Vertical Motion Analysis
The vertical motion is influenced by gravity. The vertical component of the velocity after time \( t \) is given by:
\[ V_y = u_y - g t \]
where \( g \) is the acceleration due to gravity (approximately \( 10 \, \text{m/s}^2 \)).

### Step 5: Apply the Conditions at 2 Seconds
At \( t = 2 \) seconds:
\[ V_y = u_y - g \cdot 2 \]
Since \( V_x = V_y \) and \( V_y = u_x \):
\[ u_x = u_y - 20 \]

### Step 6: Analyze the Motion at 4 Seconds
At \( t = 4 \) seconds, the particle moves horizontally, which means that the vertical component of the velocity is zero:
\[ V_y = u_y - g \cdot 4 = 0 \]
From this, we can find \( u_y \):
\[ u_y = g \cdot 4 = 10 \cdot 4 = 40 \, \text{m/s} \]

### Step 7: Substitute \( u_y \) into the Equation
Now, substitute \( u_y \) back into the equation from Step 5:
\[ u_x = 40 - 20 = 20 \, \text{m/s} \]

### Step 8: Calculate the Initial Velocity
The initial velocity \( u \) can be calculated using the Pythagorean theorem:
\[ u = \sqrt{u_x^2 + u_y^2} = \sqrt{(20)^2 + (40)^2} = \sqrt{400 + 1600} = \sqrt{2000} = 20\sqrt{5} \, \text{m/s} \]

### Final Answer
The initial velocity of the particle at the time of projection is:
\[ u = 20\sqrt{5} \, \text{m/s} \]

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