A man throws ball with the same speed vertically upwards one after the other at an interval of 2 seconds. What should be the speed of the throw so that more than two ball are in the sky at any time (Given `g=10(m)/(2^2)`)

Options

A. Any speed less than `19.6 m//s`
B. Only with speed `19.6 m//s`
C. More than `19.6 m//s`
D. Al least `9.8 m//s`

Answer

To solve the problem, we need to determine the speed at which a man should throw a ball vertically upwards so that more than two balls are in the sky at any given time.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- The man throws a ball every 2 seconds.
- We want to find the speed (u) such that there are more than two balls in the air at the same time.

2. **Time of Flight**:
- The total time of flight (T) for a ball thrown upwards can be calculated using the formula:
\[ T = \frac{2u}{g} \]
- Here, \( g \) is the acceleration due to gravity. Given \( g = 10 \, \text{m/s}^2 \).

3. **Condition for More than Two Balls**:
- For there to be more than two balls in the sky, the time of flight of the ball must be greater than the time interval between throws.
- Since the man throws a ball every 2 seconds, we need:
\[ T > 4 \, \text{seconds} \]
- This is because, at the moment the first ball is thrown, the second ball is thrown after 2 seconds, and the third ball will be in the air for 4 seconds.

4. **Setting Up the Equation**:
- From the time of flight formula:
\[ \frac{2u}{g} > 4 \]
- Substituting \( g = 10 \):
\[ \frac{2u}{10} > 4 \]

5. **Solving for u**:
- Rearranging the inequality:
\[ 2u > 40 \]
\[ u > 20 \, \text{m/s} \]

6. **Conclusion**:
- Therefore, the speed of the throw must be greater than 20 m/s for more than two balls to be in the sky at any time.

### Final Answer:
The speed of the throw should be greater than 20 m/s.

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