A Body is drops from a 100m high cliff and the same time another body is thrown from the ground with 25m/s velocity in upward direction. Where the two will meet?

Options

A. 50m
B. 40m
C. 20m
D. 10m

Answer

To solve the problem of where the two bodies meet, we can follow these steps:

### Step 1: Understand the motion of both bodies
- **Body A** is dropped from a height of 100 m with an initial velocity \( u_1 = 0 \, \text{m/s} \).
- **Body B** is thrown upwards from the ground with an initial velocity \( u_2 = 25 \, \text{m/s} \).

### Step 2: Write the equations of motion for both bodies
- For **Body A** (falling down):
\[ h_1 = u_1 t + \frac{1}{2} g t^2 \]
Since \( u_1 = 0 \):
\[ h_1 = \frac{1}{2} g t^2 \]
where \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity).

- For **Body B** (moving up):
\[ h_2 = u_2 t - \frac{1}{2} g t^2 \]
Substituting \( u_2 = 25 \, \text{m/s} \):
\[ h_2 = 25t - \frac{1}{2} g t^2 \]

### Step 3: Set up the equation for when the two bodies meet
The total distance covered by both bodies when they meet should equal the height of the cliff:
\[ h_1 + h_2 = 100 \, \text{m} \]
Substituting the expressions for \( h_1 \) and \( h_2 \):
\[ \frac{1}{2} g t^2 + (25t - \frac{1}{2} g t^2) = 100 \]
This simplifies to:
\[ 25t = 100 \]

### Step 4: Solve for time \( t \)
\[ t = \frac{100}{25} = 4 \, \text{s} \]

### Step 5: Calculate the height at which they meet
Now, we can find the height at which Body A has fallen after 4 seconds:
\[ h_1 = \frac{1}{2} g t^2 = \frac{1}{2} \times 10 \times (4)^2 = \frac{1}{2} \times 10 \times 16 = 80 \, \text{m} \]

### Step 6: Find the height of Body B
Since the total height is 100 m, the height at which Body B reaches is:
\[ h_2 = 100 - h_1 = 100 - 80 = 20 \, \text{m} \]

### Conclusion
The two bodies will meet at a height of **20 meters** from the ground.

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