A stone is dropped freely while another thrown vertically downward with an initial velocity of 4 m/s from the same point simultaneously. The distance of separation between them will become 30 m after a time of

Options

A. 8 sec
B. 10 sec
C. 15 sec
D. 7.5 sec

Answer

To solve the problem, we need to analyze the motion of both stones and find the time at which the distance between them becomes 30 meters.

### Step-by-Step Solution:

1. **Identify the motions of the two stones:**
- The first stone is dropped freely, which means it has an initial velocity of \( u_1 = 0 \, \text{m/s} \) and it accelerates downward due to gravity \( g \).
- The second stone is thrown downward with an initial velocity of \( u_2 = 4 \, \text{m/s} \) and also accelerates downward due to gravity \( g \).

2. **Write the equations for the distances traveled by each stone:**
- For the first stone (dropped freely):
\[ s_1 = u_1 t + \frac{1}{2} g t^2 = 0 \cdot t + \frac{1}{2} g t^2 = \frac{1}{2} g t^2 \]
- For the second stone (thrown downward):
\[ s_2 = u_2 t + \frac{1}{2} g t^2 = 4t + \frac{1}{2} g t^2 \]

3. **Set up the equation for the distance of separation:**
- The distance of separation between the two stones is given as:
\[ s_2 - s_1 = 30 \, \text{m} \]
- Substituting the expressions for \( s_1 \) and \( s_2 \):
\[ (4t + \frac{1}{2} g t^2) - \left(\frac{1}{2} g t^2\right) = 30 \]
- This simplifies to:
\[ 4t = 30 \]

4. **Solve for time \( t \):**
- Rearranging the equation gives:
\[ t = \frac{30}{4} = 7.5 \, \text{s} \]

5. **Conclusion:**
- The distance of separation between the two stones will become 30 meters after a time of \( t = 7.5 \, \text{s} \).

### Final Answer:
The time after which the distance of separation between the two stones becomes 30 meters is **7.5 seconds**.

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