NEET Physics Question - Kinematics - Motion Under Gravity

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:

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, \(2u/g > 2 \times 2 = 4\text{ s}\)).
\(u > 2g = 2 \times 9.8 = 19.6\text{ m/s}\).

Rankers Physics

Multiple Balls in Air

Solution:

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