A particle is projected in the `X-Y` plane. `2 sec` after projection the velocity of the particle makes an angle `45^(@)` with the `X-`axis. `4sec` after projection, it moves horizontally. Find the velocity of projection (use `g = 10 ms^(-2))`.

Options

A. `20sqrt(5) ms^-1`
B. `10sqrt(5) ms^-1`
C. `5sqrt(5) ms^-1`
D. `2sqrt(5) ms^-1`

Answer

To solve the problem step by step, we will analyze the motion of the particle projected in the X-Y plane.

### Step 1: Understanding the Motion
The particle is projected at an angle with respect to the horizontal (X-axis). We know that:
- After 2 seconds, the velocity makes an angle of 45° with the X-axis.
- After 4 seconds, the particle moves horizontally, meaning its vertical velocity component is zero at that time.

### Step 2: Analyzing the Vertical Motion
At the highest point of the trajectory (4 seconds after projection), the vertical component of the velocity (Vy) is zero. We can use the equation of motion for vertical velocity:
\[ V_y = U_y - g t \]
At \( t = 4 \) seconds:
\[ 0 = U_y - g \cdot 4 \]
Given \( g = 10 \, \text{m/s}^2 \):
\[ 0 = U_y - 40 \]
Thus, we find:
\[ U_y = 40 \, \text{m/s} \]

### Step 3: Analyzing the Motion at 2 Seconds
At \( t = 2 \) seconds, the velocity makes an angle of 45° with the X-axis. This means:
\[ V_x = V_y \]
Using the vertical motion equation again at \( t = 2 \) seconds:
\[ V_y = U_y - g \cdot 2 \]
Substituting \( U_y = 40 \):
\[ V_y = 40 - 20 = 20 \, \text{m/s} \]
Since \( V_x = V_y \):
\[ V_x = 20 \, \text{m/s} \]

### Step 4: Finding the Initial Velocity Components
Now we have:
- \( U_y = 40 \, \text{m/s} \)
- \( U_x = V_x = 20 \, \text{m/s} \)

### Step 5: Calculating the Magnitude of the Initial Velocity
The magnitude of the initial velocity \( U \) can be calculated using the Pythagorean theorem:
\[ U = \sqrt{U_x^2 + U_y^2} \]
Substituting the values:
\[ U = \sqrt{(20)^2 + (40)^2} \]
\[ U = \sqrt{400 + 1600} \]
\[ U = \sqrt{2000} \]
\[ U = 20\sqrt{5} \, \text{m/s} \]

### Final Answer
The velocity of projection is:
\[ U = 20\sqrt{5} \, \text{m/s} \]