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, we need to analyze the motion of both particles. Let's break it down step by step.

### Step 1: Define the motion of the first particle (A)
The first particle is projected upwards from point A with an initial velocity \( u_A = 50 \, \text{m/s} \). The distance traveled by this particle after time \( t \) can be given by the equation of motion:
\[ S_1 = u_A t - \frac{1}{2} g t^2 \]
Substituting the values:
\[ S_1 = 50t - \frac{1}{2} \times 10 \times t^2 = 50t - 5t^2 \]

### Step 2: Define the motion of the second particle (B)
The second particle is dropped from point B, which is 200 m above point A. Since it is dropped, its initial velocity \( u_B = 0 \). The distance traveled by this particle after time \( t \) is given by:
\[ S_2 = \frac{1}{2} g t^2 \]
Substituting the values:
\[ S_2 = \frac{1}{2} \times 10 \times t^2 = 5t^2 \]

### Step 3: Set up the equation for the total distance
When the two particles cross each other, the sum of the distances they have traveled will equal the initial distance between them, which is 200 m. Therefore, we can write:
\[ S_1 + S_2 = 200 \]
Substituting the expressions for \( S_1 \) and \( S_2 \):
\[ (50t - 5t^2) + (5t^2) = 200 \]
This simplifies to:
\[ 50t = 200 \]

### Step 4: Solve for time \( t \)
Now, we can solve for \( t \):
\[ t = \frac{200}{50} = 4 \, \text{s} \]

### Conclusion
The two particles cross each other after **4 seconds**.

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