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. at least 0.8 m/s
B. any speed less than 20 m/s
C. only with speed 20 m/s
D. more than 20 m/s

Explanation

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 time. Given that the interval between throws is 2 seconds and the acceleration due to gravity \( g = 10 \, \text{m/s}^2 \), we can follow these steps:

### Step 1: Understand the time of flight for a ball thrown upwards
The time of flight \( T \) for a ball thrown vertically upwards can be calculated using the formula:
\[ T = \frac{2U}{g} \]
where \( U \) is the initial speed of the ball and \( g \) is the acceleration due to gravity.

### Step 2: Determine the condition for more than two balls in the sky
For more than two balls to be in the sky at the same time, the time of flight of the first ball must be greater than the time taken to throw two additional balls. Since the man throws a ball every 2 seconds, the time taken to throw two balls is:
\[ 2 \, \text{s} + 2 \, \text{s} = 4 \, \text{s} \]
Thus, we need:
\[ T > 4 \, \text{s} \]

### Step 3: Substitute the time of flight into the inequality
Substituting the expression for \( T \) into the inequality gives:
\[ \frac{2U}{g} > 4 \]

### Step 4: Rearrange the inequality to solve for \( U \)
Multiplying both sides by \( g \) (which is positive) and then dividing by 2 gives:
\[ U > \frac{4g}{2} \]
\[ U > 2g \]

### Step 5: Substitute the value of \( g \)
Given \( g = 10 \, \text{m/s}^2 \), we substitute this value in:
\[ U > 2 \times 10 \]
\[ U > 20 \, \text{m/s} \]

### Conclusion
The speed at which the man should throw the ball must be greater than 20 m/s to ensure that more than two balls are in the sky at the same time.