Question 822729
The volume of a cone can be calculated as {{{pi*R^2*H/3}}} ,
where {{{R}}} is the radius of the base and {{{H}}} is the height of the cone.
Inside that conical container,
there is a cone of water at first,
with air above it.
When the container is inverted,
there is a cone of air on top,
and water below it.
All the cones are similar.
The radius of the conical container, {{{120cm/2=60cm}}} is
{{{60cm/"140 cm"=3/7}}} of the height.
{{{R/H=3/7}}} <---> {{{R=3H/7}}}
The ratio is the same for all the other similar cones.
Their volume is
{{{(1/3)*pi*R^2*H=(1/3)*pi*(3H/7)^2*H=(1/3)*pi*(9H^2/49)*H=pi*3*H^3/49}}}
The volume of the conical container, in cubic centimeters, is
{{{pi*3*140^3/49=168000pi=about527788}}} .
The volume of water in the container, in cubic centimeters, is
{{{pi*3*60^3/49=648000pi/49=about41546}}} .
The volume of air in the container, in cubic centimeters, is
{{{168000pi-648000pi/49=7584000pi/49=about486242}}} .
We can find the height, {{{H}}}, of the cone of air, in cm:
{{{pi*3*H^3/49=7584000pi/49}}}
{{{3*H^3=7584000}}}
{{{H^3=7584000/3}}}
{{{H^3=2528000}}}
{{{H=root(3,2528000)=about136.2}}}
Once the cone is standing on its base, there is a cone of air inside the tip of the container that has a height of {{{136.2cm}}}  .
If {{{136.2cm}}} at the tip of the cone are air, the remaining {{{highlight(h=3.8cm)}}} at the base of the cone are full of water.
{{{140cm-136.2cm=3.8cm}}}