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 step by step, we will analyze the motion of both bodies and calculate the difference in their heights after 3 seconds.

### Step 1: Understand the motion of both bodies
- **Body A**: This body is dropped from a height \( h \) with an initial velocity \( u_A = 0 \) m/s.
- **Body B**: This body is thrown downward with an initial velocity \( u_B = 5 \) m/s.

### Step 2: Calculate the distance fallen by Body A after 3 seconds
The distance fallen by an object under free fall can be calculated using the equation of motion:
\[ s = ut + \frac{1}{2}gt^2 \]
For Body A:
- \( u_A = 0 \) m/s (initial velocity)
- \( g = 9.81 \) m/s² (acceleration due to gravity)
- \( t = 3 \) s

Substituting the values:
\[ s_A = 0 \cdot 3 + \frac{1}{2} \cdot 9.81 \cdot (3^2) \]
\[ s_A = \frac{1}{2} \cdot 9.81 \cdot 9 \]
\[ s_A = \frac{1}{2} \cdot 88.29 \]
\[ s_A = 44.145 \text{ m} \]

### Step 3: Calculate the distance fallen by Body B after 3 seconds
For Body B:
- \( u_B = 5 \) m/s (initial velocity)
- \( g = 9.81 \) m/s² (acceleration due to gravity)
- \( t = 3 \) s

Using the same equation of motion:
\[ s_B = u_B t + \frac{1}{2}gt^2 \]
Substituting the values:
\[ s_B = 5 \cdot 3 + \frac{1}{2} \cdot 9.81 \cdot (3^2) \]
\[ s_B = 15 + \frac{1}{2} \cdot 9.81 \cdot 9 \]
\[ s_B = 15 + 44.145 \]
\[ s_B = 59.145 \text{ m} \]

### Step 4: Calculate the difference in heights after 3 seconds
The difference in height \( \Delta h \) between the two bodies after 3 seconds is given by:
\[ \Delta h = s_B - s_A \]
Substituting the distances calculated:
\[ \Delta h = 59.145 - 44.145 \]
\[ \Delta h = 15 \text{ m} \]

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

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