From a tower of height `40 m`, two bodies are simultaneously projected horizontally in opposite direction, with velocities `2 m s^-1 and 8 ms^-1`. respectively.
The horizontal distance between two bodies, when their velocity are perpendicular to each other, is.

Options

A. 1 m
B. 0.5 m
C. 2 m
D. 4 m

Explanation

To solve the problem, we need to find the horizontal distance between two bodies projected horizontally from a height of 40 m when their velocities become perpendicular to each other. Let's break it down step by step.

### Step 1: Identify the initial conditions
- Height of the tower (h) = 40 m
- Velocity of body A (v_A) = 2 m/s (to the left)
- Velocity of body B (v_B) = 8 m/s (to the right)

### Step 2: Determine the vertical motion
Since both bodies are projected horizontally, their initial vertical velocities are zero. The vertical motion of both bodies is influenced by gravity. The vertical velocity (v_y) at any time (t) can be calculated using the formula:
\[ v_y = g \cdot t \]
where \( g \) is the acceleration due to gravity (approximately \( 10 \, m/s^2 \)).

### Step 3: Calculate the time taken to fall 40 m
Using the equation of motion for vertical displacement:
\[ h = \frac{1}{2} g t^2 \]
we can rearrange this to find the time \( t \) it takes for the bodies to fall 40 m:
\[ 40 = \frac{1}{2} \cdot 10 \cdot t^2 \]
\[ 40 = 5t^2 \]
\[ t^2 = \frac{40}{5} = 8 \]
\[ t = \sqrt{8} = 2\sqrt{2} \, \text{s} \]

### Step 4: Write the expressions for the velocities
The velocities of the two bodies at time \( t \) can be expressed as:
- Velocity of body A:
\[ \vec{v_A} = -2 \hat{i} - gt \hat{j} \]
- Velocity of body B:
\[ \vec{v_B} = 8 \hat{i} - gt \hat{j} \]

### Step 5: Determine when the velocities are perpendicular
The velocities are perpendicular when their dot product is zero:
\[ \vec{v_A} \cdot \vec{v_B} = 0 \]
Calculating the dot product:
\[ (-2 \hat{i} - gt \hat{j}) \cdot (8 \hat{i} - gt \hat{j}) = -16 + g^2 t^2 = 0 \]
This simplifies to:
\[ g^2 t^2 = 16 \]
Substituting \( g = 10 \):
\[ 100 t^2 = 16 \]
\[ t^2 = \frac{16}{100} = 0.16 \]
\[ t = 0.4 \, \text{s} \]

### Step 6: Calculate the horizontal distance
Now, we can calculate the horizontal distance traveled by both bodies in this time:
- Distance traveled by body A:
\[ d_A = v_A \cdot t = 2 \cdot 0.4 = 0.8 \, \text{m} \]
- Distance traveled by body B:
\[ d_B = v_B \cdot t = 8 \cdot 0.4 = 3.2 \, \text{m} \]

### Step 7: Find the total horizontal distance between the two bodies
The total horizontal distance between the two bodies when their velocities are perpendicular is:
\[ d_{total} = d_A + d_B = 0.8 + 3.2 = 4 \, \text{m} \]

### Final Answer
The horizontal distance between the two bodies when their velocities are perpendicular to each other is **4 meters**.

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