Q:

A body is projected from ground with some angle to the horizontal, how the angular momentum about the initial position of the motion increases? 

A body is projected from ground with some angle to the horizontal, how the angular momentum about the initial position of the motion increases? 

I have tried to solve this problem using parallel axis theorem, so that when the distance between the object and initial point increases, the moment of inertia increases and hence therefore angular momentum increases.  But I am not sure if this explanation is correct or not. 

A:

To see this conceptually, fire an object horizontally from the top of a cliff.  
It’s moving straight away from the launch point at first: zero angular momentum about that point.  
Gravity will act downward.  That’s not through the origin (because it acts where the body is instantaneously), so that’s a non-zero torque: angular momentum changes! 
The path changes to no longer be straight away from the launch point.  Rather, it has an increasing perpendicular component of velocity, around the origin: that’s the angular momentum created by gravity.  
Now mathematically:  $L(t)$ is $r(t) \times v(t)$.  You can use your knowledge of projectile motion to write $r(t)$ and $v(t)$ in component form ($x(t)$, etc) leaving launch angle and speed as variables.   Then write out the cross product in components, which is easy in this case:
$L(t) = x(t)v_y(t) - y(t)v_x(t)$
You'll get something with clear time variance.