A stone thrown upwards attains a maximum height of 19.6m. Find the velocity with which it was thrown.

We need to find the initial velocity with which the stone was thrown in the air.
It can be calculated with the help of kinematical equations.

Given: 
Distance covered or height, h= 19.6m
Acceleration due to gravity, g= 9.8m/s²
When a stone reaches the maximum height, its velocity becomes zero before falling down.
Final velocity, v=0

To find: 
Initial velocity, u=0

Formula:
$v^2=u^2+2as$

Solution:
For this problem, we can modify the kinematical equation as:
$v^2=u^2+2gh$

$0=u^2 +2gh$ 
$u^2=-2gh$
Since initial velocity and acceleration due to gravity are in opposite directions, the negative sign will get canceled.

$\text{u}=\sqrt{\text{2gh}}$

$=\sqrt{2\times 9.8\times 19.6}$

$\text{u}=19.6\text{m/s}$