A particle is projected from a point A verically upwards with a speed of `50 ms ^(-1)` and another is dropped simulataneously from B which is 200 m veritically above A. They cross each other after [Given: `g = 10 ms ^(-2)].`

Options

A. 8 s
B. 6 s
C. 4 s
D. 5 s

Answer

To solve the problem step by step, we will analyze the motion of both particles and find the time at which they cross each other.

### Step-by-Step Solution

1. **Identify the Initial Conditions:**
- The first particle (let's call it Particle 1) is projected upwards from point A with an initial velocity \( u_1 = 50 \, \text{m/s} \).
- The second particle (Particle 2) is dropped from point B, which is 200 m above A, with an initial velocity \( u_2 = 0 \, \text{m/s} \).

2. **Define the Motion Equations:**
- For Particle 1 (moving upwards):
\[ S_1 = u_1 t - \frac{1}{2} g t^2 \]
Substituting \( u_1 = 50 \, \text{m/s} \) and \( g = 10 \, \text{m/s}^2 \):
\[ S_1 = 50t - 5t^2 \]

- For Particle 2 (moving downwards):
\[ S_2 = u_2 t + \frac{1}{2} g t^2 \]
Substituting \( u_2 = 0 \, \text{m/s} \) and \( g = 10 \, \text{m/s}^2 \):
\[ S_2 = 0 + 5t^2 = 5t^2 \]

3. **Set Up the Equation for the Total Distance:**
- The total distance between the two particles when they meet is 200 m:
\[ S_1 + S_2 = 200 \]
- Substituting the expressions for \( S_1 \) and \( S_2 \):
\[ (50t - 5t^2) + 5t^2 = 200 \]
- Simplifying this gives:
\[ 50t = 200 \]

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

5. **Conclusion:**
- The two particles cross each other after \( t = 4 \) seconds.