A body is dropped from a certain height h (h is very large) and second bodyis thrown downward with velocity of 5 m/s simultaneouly. What will be difference in heights of the two bodies after 3 s?

Options

A. 5 m
B. 10 m
C. 15 m
D. 20 m

Answer

To solve the problem, we need to determine the heights fallen by both bodies after 3 seconds and then find the difference in their heights.

### Step-by-Step Solution:

1. **Identify the Initial Conditions:**
- For the first body (dropped from height \( h \)):
- Initial velocity \( u_1 = 0 \, \text{m/s} \)
- For the second body (thrown downward):
- Initial velocity \( u_2 = 5 \, \text{m/s} \)

2. **Use the Equation of Motion for the First Body:**
The distance fallen by the first body after time \( t \) can be calculated using the equation:
\[ s_1 = u_1 t + \frac{1}{2} g t^2 \]
Substituting the values:
\[ s_1 = 0 \cdot 3 + \frac{1}{2} g (3)^2 = \frac{1}{2} g \cdot 9 = \frac{9g}{2} \, \text{meters} \]

3. **Use the Equation of Motion for the Second Body:**
The distance fallen by the second body after time \( t \) is given by:
\[ s_2 = u_2 t + \frac{1}{2} g t^2 \]
Substituting the values:
\[ s_2 = 5 \cdot 3 + \frac{1}{2} g (3)^2 = 15 + \frac{1}{2} g \cdot 9 = 15 + \frac{9g}{2} \, \text{meters} \]

4. **Calculate the Difference in Heights:**
The difference in heights \( \Delta h \) between the two bodies after 3 seconds is:
\[ \Delta h = s_2 - s_1 \]
Substituting the values we calculated:
\[ \Delta h = \left(15 + \frac{9g}{2}\right) - \left(\frac{9g}{2}\right) \]
Simplifying this gives:
\[ \Delta h = 15 \, \text{meters} \]

### Final Answer:
The difference in heights of the two bodies after 3 seconds is **15 meters**.