A light body is projected with a velocity `(10hat(i)+20 hat(j)+20 hat(k)) ms^(-1)`. Wind blows along X-axis with an acceleration of `2.5 ms^(2)`. If Y-axis is vertical then the speed of particle after 2 second will be `(g=10 ms^(2))`

Options

A. `25 ms^(-1)`
B. `10sqrt(5) ms^(-1)`
C. `30 ms^(-1)`
D. none of these

Answer

To solve the problem step by step, we will break down the motion of the particle in three dimensions (x, y, and z) and calculate the speed after 2 seconds.

### Given:
- Initial velocity \( \mathbf{u} = 10 \hat{i} + 20 \hat{j} + 20 \hat{k} \) m/s
- Acceleration due to wind along x-axis \( a_x = 2.5 \) m/s²
- Acceleration due to gravity (acting downwards along y-axis) \( g = 10 \) m/s²
- Time \( t = 2 \) seconds

### Step 1: Calculate the velocity in the x-direction after 2 seconds.
The formula for velocity under constant acceleration is:
\[ v_x = u_x + a_x t \]
Where:
- \( u_x = 10 \) m/s (initial velocity in x-direction)
- \( a_x = 2.5 \) m/s² (acceleration in x-direction)
- \( t = 2 \) s

Substituting the values:
\[ v_x = 10 + (2.5 \times 2) = 10 + 5 = 15 \text{ m/s} \]

### Step 2: Calculate the velocity in the y-direction after 2 seconds.
The formula for velocity in the y-direction is:
\[ v_y = u_y - g t \]
Where:
- \( u_y = 20 \) m/s (initial velocity in y-direction)
- \( g = 10 \) m/s² (acceleration due to gravity)
- \( t = 2 \) s

Substituting the values:
\[ v_y = 20 - (10 \times 2) = 20 - 20 = 0 \text{ m/s} \]

### Step 3: The velocity in the z-direction remains constant.
Since there is no acceleration in the z-direction, the velocity remains the same:
\[ v_z = u_z = 20 \text{ m/s} \]

### Step 4: Combine the velocities to find the resultant velocity vector.
The resultant velocity vector \( \mathbf{v} \) after 2 seconds is:
\[ \mathbf{v} = v_x \hat{i} + v_y \hat{j} + v_z \hat{k} = 15 \hat{i} + 0 \hat{j} + 20 \hat{k} \text{ m/s} \]

### Step 5: Calculate the magnitude of the resultant velocity.
The magnitude of the velocity \( |\mathbf{v}| \) is given by:
\[ |\mathbf{v}| = \sqrt{v_x^2 + v_y^2 + v_z^2} \]
Substituting the values:
\[ |\mathbf{v}| = \sqrt{15^2 + 0^2 + 20^2} = \sqrt{225 + 0 + 400} = \sqrt{625} = 25 \text{ m/s} \]

### Final Answer:
The speed of the particle after 2 seconds is \( 25 \) m/s.

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