Two bodies are dropped from two different building, one at a height 200m and another at a height of 150m above the ground. What will be difference in height of the bodies after they have fallen for 2s? `(g=10m//s^(2))`

Answer

To solve the problem of finding the difference in height between two bodies dropped from different heights after 2 seconds, we can follow these steps:

### Step 1: Identify the initial conditions
- Body 1 is dropped from a height of 200 meters.
- Body 2 is dropped from a height of 150 meters.
- Both bodies are dropped from rest, so their initial velocity (u) is 0 m/s.
- The acceleration due to gravity (g) is given as 10 m/s².

### Step 2: Calculate the distance fallen by each body after 2 seconds
We will use the second equation of motion:
\[ s = ut + \frac{1}{2}gt^2 \]
Where:
- \( s \) = distance fallen
- \( u \) = initial velocity (0 m/s)
- \( g \) = acceleration due to gravity (10 m/s²)
- \( t \) = time (2 seconds)

For both bodies, since \( u = 0 \):
\[ s = 0 \cdot t + \frac{1}{2} \cdot 10 \cdot (2)^2 \]
\[ s = \frac{1}{2} \cdot 10 \cdot 4 \]
\[ s = 20 \text{ meters} \]

### Step 3: Calculate the height of each body above the ground after 2 seconds
- For Body 1 (initial height = 200 m):
\[ \text{Height after 2 seconds} = 200 \text{ m} - 20 \text{ m} = 180 \text{ m} \]

- For Body 2 (initial height = 150 m):
\[ \text{Height after 2 seconds} = 150 \text{ m} - 20 \text{ m} = 130 \text{ m} \]

### Step 4: Calculate the difference in height between the two bodies
Now we find the difference in height between the two bodies after 2 seconds:
\[ \text{Difference in height} = \text{Height of Body 1} - \text{Height of Body 2} \]
\[ \text{Difference in height} = 180 \text{ m} - 130 \text{ m} = 50 \text{ m} \]

### Final Answer
The difference in height between the two bodies after they have fallen for 2 seconds is **50 meters**.

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