Question

A man throws balls with the same speed vertically upwards one after the other at an interval of $2s$. What should be the speed of the throw so that more than two balls are in the sly at any time? (Given $g = 9.8\, m/s^2$)

• 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. At least 9.8 m/s

Answer 1

The Correct Option is C

To determine the speed with which the balls must be thrown such that more than two balls are in the air at any given time, we can analyze the physics behind the projectile motion of each ball. 

When a ball is thrown upwards with an initial speed \(v\), it reaches a maximum height where its velocity becomes zero, and then it falls back down. The total time \(T\) it stays in the air (time to go up and come down) can be given by the formula:

\(T = \frac{2v}{g}\)

where \(g = 9.8\, m/s^2\) is the acceleration due to gravity.

If balls are thrown every 2 seconds, for more than two balls to be in the air simultaneously, the time \(T\) must be greater than 4 seconds (since if it were exactly 4 seconds, only two balls would be in the air at the same time).

Thus, setting \(\frac{2v}{g} > 4\), we solve for \(v\):

\(\frac{2v}{9.8} > 4\)

Multiplying both sides by 9.8:

\(2v > 39.2\)

Dividing both sides by 2 gives:

\(v > 19.6\)

Thus, the speed of the throw must be more than \(19.6\, m/s\) for more than two balls to be in the air at any time.

Conclusion: The correct answer is "More than 19.6 m/s".

This ensures that each ball remains airborne long enough to overlap the subsequent throws, ensuring more than two are in the air simultaneously.