Two skaters P and Q are skating towards each other. Skater P throws a ball towards W every 5 s such that it always leaves her hand with speed 2 `ms^(-1)` with respect to the ground. Consider two cases:
(I) P runs with speed 1 `ms^(-1)` towards Q while Q remains stationary
(II) Q runs with speed 1 `ms^(-1)` towards P while P remains stationary.
Note that irrespective of speed of P, ball always leaves P's hand with speed `2 ms^(-1)` with respect to the ground. Ignore gravity. Balls will be received by Q .

Options

A. one every 2.5 s in case (I) and one every 3.3 s in case (II)
B. one every 2 s in case (I) and one every 4 s in case (II)
C. one every 3.3 s in case (I) and one every 2.5 s in case (II)
D. one every 2.5 s in case (I) and one every 2.5 s in case (II)

Explanation

To solve the problem, we will analyze both cases step by step.

### Case I: Skater P runs with speed 1 m/s towards Q while Q remains stationary.

1. **Initial Setup**:
- Let the initial distance between P and Q be \( x \).
- Skater P throws the ball every 5 seconds with a speed of 2 m/s relative to the ground.

2. **Time for the First Ball**:
- The first ball is thrown at \( t = 0 \) seconds.
- The time taken for the ball to reach Q can be calculated using the formula:
\[ T_1 = \frac{x}{\text{speed of the ball}} = \frac{x}{2} \text{ seconds} \]

3. **Position of P after 5 seconds**:
- After 5 seconds, P will have moved towards Q:
\[ \text{Distance covered by P} = \text{speed} \times \text{time} = 1 \text{ m/s} \times 5 \text{ s} = 5 \text{ m} \]
- The new distance between P and Q after 5 seconds will be:
\[ x - 5 \text{ m} \]

4. **Time for the Second Ball**:
- The second ball is thrown at \( t = 5 \) seconds.
- The time taken for the second ball to reach Q is:
\[ T_2 = \frac{x - 5}{2} \text{ seconds} \]

5. **Total Time for Each Ball**:
- The total time for the first ball to reach Q is:
\[ T_1 = \frac{x}{2} \]
- The total time for the second ball to reach Q is:
\[ T_2 = 5 + \frac{x - 5}{2} \]

6. **Finding the Time Difference**:
- The time difference between when Q receives the first and second ball is:
\[ T_2 - T_1 = \left(5 + \frac{x - 5}{2}\right) - \frac{x}{2} \]
- Simplifying this gives:
\[ T_2 - T_1 = 5 + \frac{x - 5 - x}{2} = 5 - \frac{5}{2} = 5 - 2.5 = 2.5 \text{ seconds} \]

### Case II: Q runs with speed 1 m/s towards P while P remains stationary.

1. **Initial Setup**:
- Again, let the initial distance between P and Q be \( x \).
- The speed of the ball is still 2 m/s.

2. **Time for the First Ball**:
- The first ball is thrown at \( t = 0 \) seconds.
- The effective speed of the ball relative to Q is:
\[ \text{Speed of the ball relative to Q} = 2 + 1 = 3 \text{ m/s} \]
- The time taken for the first ball to reach Q is:
\[ T_1 = \frac{x}{3} \text{ seconds} \]

3. **Position of Q after 5 seconds**:
- After 5 seconds, Q will have moved towards P:
\[ \text{Distance covered by Q} = 1 \text{ m/s} \times 5 \text{ s} = 5 \text{ m} \]
- The new distance between P and Q after 5 seconds will be:
\[ x - 5 \text{ m} \]

4. **Time for the Second Ball**:
- The second ball is thrown at \( t = 5 \) seconds.
- The time taken for the second ball to reach Q is:
\[ T_2 = \frac{x - 5}{3} \text{ seconds} \]

5. **Total Time for Each Ball**:
- The total time for the first ball to reach Q is:
\[ T_1 = \frac{x}{3} \]
- The total time for the second ball to reach Q is:
\[ T_2 = 5 + \frac{x - 5}{3} \]

6. **Finding the Time Difference**:
- The time difference between when Q receives the first and second ball is:
\[ T_2 - T_1 = \left(5 + \frac{x - 5}{3}\right) - \frac{x}{3} \]
- Simplifying this gives:
\[ T_2 - T_1 = 5 + \frac{x - 5 - x}{3} = 5 - \frac{5}{3} = 5 - 1.67 = \frac{15 - 5}{3} = \frac{10}{3} \approx 3.33 \text{ seconds} \]

### Summary of Results:
- In Case I, Q receives the first ball after **2.5 seconds**.
- In Case II, Q receives the first ball after **3.33 seconds**.