A man throws balls with the same speed vertically upwards one after the other at an interval of 2 seconds. What should be the speed of the throw so that more than two balls are in the sky at any time (Given 9.8m/s²)

To find the required speed of projection, we can analyze the motion of the balls step-by-step.

Let $u$ be the initial speed with which each ball is thrown vertically upwards, and $g$ be the acceleration due to gravity, given as $g=9.8{\mathrm{m/s}}^2$.

First, let's determine the total time a single ball remains in the sky (its time of flight, $T$). The time taken to reach the maximum height is:
$t=\frac{u}{g}$
Since the time of ascent is equal to the time of descent, the total time of flight $T$ in the air is:
$T=\frac{2u}{g}$

The balls are thrown one after another at a constant time interval of $2\text{seconds}$.

For more than two balls to be in the sky at any given instant, the first ball thrown must still be in the air when the third ball is projected.
The third ball is projected at $t=4\text{seconds}$ (since the second ball is thrown at $t=2\text{seconds}$).
Therefore, the total time of flight of the first ball must be strictly greater than $4\text{seconds}$:
$T>4$

Substituting the expression for $T$ into the inequality:
$\frac{2u}{g}>4$

Solving for $u$:
$u>2g$

Now, substitute the value of $g=9.8{\mathrm{m/s}}^2$:
$u>2\times 9.8$
$u>19.6\text{m/s}$

Thus, the speed of projection must be more than 19.6 m/s.