A man throws balls with the same speed vertically upwards one after the other at an interval of 2 s what should be the speed of the throw so that more than two balls are in the sky at any time?

Hint: When an object is thrown upward it is under the negative acceleration of gravitational force so its speed value will decrease at the highest point its final velocity will be zero then again it will experience positive acceleration and again when it will strike the ground its velocity will be zero. Here it should be noted that the distance travelled by going up and going down will be the same in both cases.

Complete step by step solution:
Here in this question, it is said that balls are thrown at an interval of 2 secs and it is asked what should be the velocity that the balls should be thrown so that there are more than 2 balls at a point of time.
More than 2 means at least 3
Now, one ball is thrown then after 2 secs the second ball is thrown and then after another 2 secs the third ball is thrown. So, the time of flight of the first ball should be more than 4 secs so that all the three balls should be in the air
Now when the first ball is thrown
Lets its initial velocity be U
Final velocity will be V
The time for the flight will be T
Using laws of motion
$\Rightarrow V = U + aT$
Here acceleration is against motion so the equation becomes
$\Rightarrow V = U - gT$
$\because V = 0$
$ \Rightarrow U = gT$
$ \Rightarrow T = \dfrac{U}{g}$
The ball first goes up and then comes down so the total period be 2T and the minimum period should be more than 4 secs
$\Rightarrow 4 > 2\dfrac{U}{g}$
g is given $10m{\sec ^{ - 1}}$
\[ \Rightarrow U > \dfrac{{40}}{2}\]
\[ \Rightarrow U > 20m{\sec ^{ - 1}}\]
Final Answer: If the balls are thrown at $20m{\sec ^{ - 1}}$ then at that speed 3 balls will be in the sky at the same time.

Note: At this speed, the first ball will be just above the ground before landing.
The second ball will be around the top point.
The third ball will be just thrown.