A particle is projected from a point A vertically upwards with a speed of `50 ms ^(-1)` and another is dropped simultaneously from B which is 200 m vertically 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 of two particles crossing each other, we can follow these steps:

### Step 1: Define the motion of both particles
- **Particle A** is projected upwards from point A with an initial velocity \( u_A = 50 \, \text{m/s} \).
- **Particle B** is dropped from point B, which is 200 m above A, with an initial velocity \( u_B = 0 \, \text{m/s} \).

### Step 2: Write the equations of motion
Using the second equation of motion \( s = ut + \frac{1}{2} a t^2 \):

1. For **Particle A** (upward motion):
\[ s_1 = u_A t - \frac{1}{2} g t^2 = 50t - \frac{1}{2} \cdot 10 t^2 = 50t - 5t^2 \]

2. For **Particle B** (downward motion):
\[ s_2 = u_B t + \frac{1}{2} g t^2 = 0 + \frac{1}{2} \cdot 10 t^2 = 5t^2 \]

### Step 3: Set up the equation for the total distance
Since the total distance between the two particles is 200 m, we can write:
\[ s_1 + s_2 = 200 \]
Substituting the expressions for \( s_1 \) and \( s_2 \):
\[ (50t - 5t^2) + 5t^2 = 200 \]

### Step 4: Simplify the equation
The equation simplifies to:
\[ 50t = 200 \]
This means:
\[ 50t = 200 \implies t = \frac{200}{50} = 4 \, \text{seconds} \]

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