Multiple Balls in Air – Rankers Physics

Motion Under Gravity: Practice Problem & Solution

A man throws ball with the same speed vertically upwards one after the other at an interval of \(2\text{ seconds}\)). What should be the speed of the throw so that more than two balls are in the sky at any time? (Given \(g = 9.8 m/s^2\) (2003)
More than \(19.6\text{ m/s}\)
At least \(9.8\text{ m/s}\)
Any speed less than \(19.6\text{ m/s}\)
Only with speed \(19.6\text{ m/s}\)

Solution Explained:

To solve this problem, we apply the core principles of Motion Under Gravity. Understanding the underlying formula is key to arriving at the correct answer below:

Concept: Time of flight for vertical motion. Let \(\Delta t = 2\text{ s}\)) be the throwing interval.
Formula: Time of flight \(T = 2u/g\).
For more than two balls to be in the air, the time of flight of each ball must be greater than twice the interval: \(T > 2\Delta t\).
So, \(2u/g > 2 \times 2 = 4\text{ s}\)).
\(u > 2g = 2 \times 9.8 = 19.6\text{ m/s}\).

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