A stone is dropped freely, while another thrown vertically downward with an initial velocity of `2 "ms"^(-1)` from the same point, simultaneously. The time required by them to have a distance of separation 22 m between them is

Options

A. 11 s
B. 5.5 s
C. 44 s
D. 22 s

Answer

To solve the problem of finding the time required for the two stones to have a separation of 22 meters, we can follow these steps:

### Step 1: Define the Variables
Let:
- Stone A is dropped freely (initial velocity \( u_A = 0 \, \text{m/s} \)).
- Stone B is thrown downward with an initial velocity \( u_B = 2 \, \text{m/s} \).
- Both stones are subjected to the same gravitational acceleration \( g = 9.81 \, \text{m/s}^2 \).

### Step 2: Write the Equations of Motion
For Stone A (dropped freely):
- The distance fallen after time \( t \) is given by:
\[ s_A = u_A t + \frac{1}{2} g t^2 = 0 + \frac{1}{2} g t^2 = \frac{1}{2} g t^2 \]

For Stone B (thrown downward):
- The distance fallen after time \( t \) is given by:
\[ s_B = u_B t + \frac{1}{2} g t^2 = 2t + \frac{1}{2} g t^2 \]

### Step 3: Calculate the Distance Between the Two Stones
The distance of separation \( d \) between Stone B and Stone A after time \( t \) is:
\[ d = s_B - s_A \]
Substituting the equations for \( s_A \) and \( s_B \):
\[ d = \left(2t + \frac{1}{2} g t^2\right) - \left(\frac{1}{2} g t^2\right) \]
This simplifies to:
\[ d = 2t \]

### Step 4: Set Up the Equation for Separation
We want the distance of separation to equal 22 meters:
\[ 2t = 22 \]

### Step 5: Solve for Time \( t \)
To find \( t \):
\[ t = \frac{22}{2} = 11 \, \text{seconds} \]

### Final Answer
The time required for the two stones to have a separation of 22 meters is **11 seconds**.

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