Step 1

We are given a buoy made from two homogeneous cones. The following data is given:

The base radius of each of the cones is $r = 1.5{\rm{ ft}}$.

The center of gravity distance of the buoy is $\overline z = 0.5{\rm{ ft}}$.

The depth of the bottom cone below the liquid is ${h_1} = 4{\rm{ ft}}$.

We are required to determine the height $h$ of the top cone.

 
Step 2

The diagram of the buoy showing different segments is given below:

Here, point $G$ represents the center of gravity of the buoy and $\overline z $ represents the distance of the center of gravity of the buoy.

 
Step 3

The volume of the top cone segment 1 is given by,

\[\begin{array}{l} {V_1} = \frac{\pi }{3}{r^2}h\\ {V_1} = \frac{\pi }{3} \times {\left( {1.5{\rm{ ft}}} \right)^2} \times h\\ {V_1} = 2.356h{\rm{ f}}{{\rm{t}}^2} \end{array}\]

The volume of the bottom cone segment 2 is given by,

\[\begin{array}{l} {V_2} = \frac{\pi }{3}{r^2}{h_1}\\ {V_2} = \frac{\pi }{3} \times {\left( {1.5{\rm{ ft}}} \right)^2} \times 4{\rm{ ft}}\\ {V_2} = 9.425{\rm{ f}}{{\rm{t}}^3} \end{array}\]

Then, the total volume of the buoy is given by,

\[\begin{array}{l} V = {V_1} + {V_2}\\ V = 2.356h{\rm{ f}}{{\rm{t}}^2} + 9.425{\rm{ f}}{{\rm{t}}^3} \end{array}\]
 
Step 4

The $z$ component of the center of gravity of the segment 1 aligned with the $z$ axis is given by,

\[{z_1} = \frac{{ - h}}{4}\]

The $z$ component of the center of gravity of the segment 2 aligned with the $z$ axis is given by,

\[\begin{array}{l} {z_2} = \frac{{{h_1}}}{4}\\ {z_2} = \frac{{4{\rm{ ft}}}}{4}\\ {z_2} = 1{\rm{ ft}} \end{array}\]
 
Step 5

Then, using the formula for the center of gravity, the center of gravity of the buoy is given by,

\[\overline z = \frac{{{z_1}{V_1} + {z_2}{V_2}}}{V}\]

On substituting the values in the above expression, we get,

\[\begin{array}{c} 0.5{\rm{ ft}} = \frac{{\left( {\frac{{ - h}}{4}} \right) \times \left( {2.356h{\rm{ f}}{{\rm{t}}^2}} \right) + \left( {1{\rm{ ft}}} \right) \times \left( {9.425{\rm{ f}}{{\rm{t}}^3}} \right)}}{{2.356h{\rm{ f}}{{\rm{t}}^2} + 9.425{\rm{ f}}{{\rm{t}}^3}}}\\ 1.178h{\rm{ f}}{{\rm{t}}^3} + 4.7125{\rm{ f}}{{\rm{t}}^4} = - 0.589{h^2}{\rm{ f}}{{\rm{t}}^2} + 9.425{\rm{ f}}{{\rm{t}}^4}\\ 0.589{h^2}{\rm{ f}}{{\rm{t}}^2} + 1.178h{\rm{ f}}{{\rm{t}}^3} - 4.7125{\rm{ f}}{{\rm{t}}^4} = 0 \end{array}\]

On solving the above quadratic equation, we get,

\[\begin{array}{c} h = \frac{{ - \left( {1.178{\rm{ f}}{{\rm{t}}^3}} \right) \pm \sqrt {{{\left( {1.178{\rm{ f}}{{\rm{t}}^3}} \right)}^2} - \left\{ {4 \times \left( {0.589{\rm{ f}}{{\rm{t}}^2}} \right) \times \left( { - 4.7125{\rm{ f}}{{\rm{t}}^4}} \right)} \right\}} }}{{2 \times \left( {0.589{\rm{ f}}{{\rm{t}}^2}} \right)}}\\ h = \frac{{ - 1.178{\rm{ f}}{{\rm{t}}^3} \pm 3.534{\rm{ f}}{{\rm{t}}^3}}}{{1.178{\rm{ f}}{{\rm{t}}^2}}}\\ h = 2{\rm{ ft}}\\ h = - 4{\rm{ ft}} \end{array}\]

Considering the positive root only, we have,

\[h = 2{\rm{ ft}}\]