Question

a man throws ball with the same speed vertically upward one after the other at an interval of 2 seconds.what should be the speed of the throws so that more than two balls are in the sky at any time?

Answer 1

When you throw the ball upwards the final velocity becomes zero therfore,
v=u+at
0= u-gt
u= 10×2=20 m/s
Hence, when you throw a ball with minimum 20m/s in a time interval of two seconds two balls can be in air simulataneously....

Answer 2

u = 19.6 m/s  

Given:

The man throws a ball with the speed which equal for the entire ball and every ball is thrown at an interval of 2 seconds. T

Solution:

The speed of the ball is calculated using first equation of motion, v = u + at.

So, the equation formed in this situation will be more than 2 balls be in air means let us assume 3 balls so the time taken by first ball in flight be more than or equal to 4 seconds, as per the situation.  

Therefore, we can say that when t = 4 seconds the third ball is thrown.  

Therefore the first ball equation be, v = u + at

Final velocity = -v, Initial velocity = -u, Acceleration due to gravity = -g

$\begin{array}{l}{-u=u-g t} \\ {u=\frac{g t}{2}} \\ {u=9.8 \times \frac{4}{2}} \\ \bold{u=19.6\ \mathrm{m} / \mathrm{s}}\end{array}$