A stone is dropped from a tower of height 200 m and at the same time another stone is projected vertically up at 50 m/s. The height at which they meet ? `[g=10 "m/s"^2]`

Options

A. 5 m
B. 100 m
C. 120 m
D. 125 m

Answer

To solve the problem of finding the height at which two stones meet, we can break down the solution into clear steps.

### Step-by-Step Solution:

1. **Understand the Problem:**
- A stone (let's call it Stone A) is dropped from a height of 200 m.
- Another stone (let's call it Stone B) is projected upwards with an initial velocity of 50 m/s.
- We need to find the height at which they meet.

2. **Define Variables:**
- Let \( h \) be the height of the tower: \( h = 200 \) m.
- Let \( x \) be the height at which they meet.
- The distance Stone A falls is \( x \) and the distance Stone B rises is \( h - x \).

3. **Equations of Motion:**
- For Stone A (dropped from rest):
\[ x = \frac{1}{2} g t^2 \]
where \( g = 10 \, \text{m/s}^2 \).
- For Stone B (projected upwards):
\[ h - x = ut - \frac{1}{2} g t^2 \]
where \( u = 50 \, \text{m/s} \).

4. **Substituting Values:**
- From the first equation for Stone A:
\[ x = \frac{1}{2} (10) t^2 = 5 t^2 \]
- Substitute \( x \) into the second equation:
\[ 200 - 5 t^2 = 50t - \frac{1}{2} (10) t^2 \]
Simplifying the second equation gives:
\[ 200 - 5 t^2 = 50t - 5t^2 \]
Rearranging gives:
\[ 200 = 50t \]

5. **Solve for Time \( t \):**
- From \( 200 = 50t \):
\[ t = \frac{200}{50} = 4 \, \text{s} \]

6. **Find the Height \( x \):**
- Substitute \( t \) back into the equation for \( x \):
\[ x = 5 t^2 = 5 (4^2) = 5 \times 16 = 80 \, \text{m} \]
- The height at which Stone A falls is \( 80 \, \text{m} \), so the height at which they meet is:
\[ h - x = 200 - 80 = 120 \, \text{m} \]

### Final Answer:
The height at which both stones meet is **120 meters**.