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 time taken for the displacement vectors of two bodies to be come perpendicular to each other is.

Options

A. 0.1 s
B. 0.2 s
C. 0.8 s
D. 0.6 s

Explanation

To solve the problem, we need to determine the time taken for the displacement vectors of two bodies projected horizontally from a tower to become perpendicular to each other. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the problem
We have two bodies projected horizontally from a height of 40 m. One body is projected with a velocity of 2 m/s to the left, and the other with a velocity of 8 m/s to the right. We need to find the time at which their displacement vectors become perpendicular.

### Step 2: Set up the coordinate system
Let’s define our coordinate system:
- The vertical direction (y-axis) is downward.
- The horizontal direction (x-axis) will be to the right.

### Step 3: Write the displacement equations
For body A (projected with 2 m/s):
- Horizontal displacement: \( x_A = -2t \) (to the left)
- Vertical displacement: \( y_A = -\frac{1}{2}gt^2 \) (downward)

For body B (projected with 8 m/s):
- Horizontal displacement: \( x_B = 8t \) (to the right)
- Vertical displacement: \( y_B = -\frac{1}{2}gt^2 \) (downward)

### Step 4: Write the displacement vectors
The displacement vectors for both bodies can be expressed as:
- \( \vec{r_A} = (-2t) \hat{i} + \left(-\frac{1}{2}gt^2\right) \hat{j} \)
- \( \vec{r_B} = (8t) \hat{i} + \left(-\frac{1}{2}gt^2\right) \hat{j} \)

### Step 5: Determine when the vectors are perpendicular
The displacement vectors are perpendicular when their dot product is zero:
\[ \vec{r_A} \cdot \vec{r_B} = 0 \]
Calculating the dot product:
\[ (-2t)(8t) + \left(-\frac{1}{2}gt^2\right)\left(-\frac{1}{2}gt^2\right) = 0 \]
This simplifies to:
\[ -16t^2 + \frac{g^2 t^4}{4} = 0 \]

### Step 6: Rearranging the equation
Rearranging gives:
\[ \frac{g^2 t^4}{4} = 16t^2 \]
Dividing both sides by \( t^2 \) (assuming \( t \neq 0 \)):
\[ \frac{g^2 t^2}{4} = 16 \]
Multiplying through by 4:
\[ g^2 t^2 = 64 \]

### Step 7: Solve for \( t \)
Taking the square root of both sides:
\[ t^2 = \frac{64}{g^2} \]
Thus:
\[ t = \frac{8}{g} \]
Given \( g \approx 10 \, \text{m/s}^2 \):
\[ t = \frac{8}{10} = 0.8 \, \text{s} \]

### Final Answer
The time taken for the displacement vectors of the two bodies to become perpendicular to each other is **0.8 seconds**.

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