Step 1

We are given the following data:

The weight of the block $A$ is ${W_{\rm{A}}} = 12\,{\rm{lb}}$.

The weight of block $B$ is ${W_{\rm{B}}} = 15\;{\rm{lb}}$.

The weight of block $C$ is ${W_{\rm{C}}} = 15\;{\rm{lb}}$.

We are asked to determine the maximum distance $A$ will fall before its motion is momentarily stopped.

 
Step 2

We will draw the free-body diagram of block system.

Here, $R$ is the reaction force and $y$ is the vertical downward distance of the block $A$.

From the above diagram, we can observed some distance as

The distance ${x_1}$ is ${x_1} = 4\;{\rm{ft}}$.

The distance ${x_2}$ is ${x_2} = 4\;{\rm{ft}}$.

 
Step 3

The formula to calculate the distance $t$ is given by,

\[t = \sqrt {{x_1}^2 + {y^2}} \]

Substitute all the known values in the above equation.

\[\begin{array}{c} t = \sqrt {{{\left( {4\;{\rm{ft}}} \right)}^2} + {y^2}} \\ = \sqrt {{y^2} + 16\;{\rm{f}}{{\rm{t}}^{\rm{2}}}} \end{array}\]
 
Step 4

The block released from rest and in the last its motion is momentarily stopped so the initial and final kinetic energy of the system would be considered as zero.

We will apply the principle of work and energy,

\[\begin{array}{c} {T_1} + \sum {{U_{{\rm{1 - 2}}}}} = {T_2}\\ \left[ {0 + \left[ {\left( {{W_{\rm{A}}}} \right)y - \left( {{W_{\rm{B}}} + {W_{\rm{C}}}} \right)\left( {t - {x_1}} \right)} \right]} \right] = 0\\ \left[ {\left( {{W_{\rm{A}}}} \right)y - \left( {{W_{\rm{B}}} + {W_{\rm{C}}}} \right)\left( {t - {x_1}} \right)} \right] = 0 \end{array}\]

Substitute all the known values in the above equation.

\[\begin{array}{c} \left[ \begin{array}{l} \left( {12\;{\rm{lb}}} \right)y\\ - \left( {15\;{\rm{lb}}\;{\rm{ + }}\;{\rm{15}}\;{\rm{lb}}} \right)\left\{ {\sqrt {{y^2} + \left( {16\;{\rm{f}}{{\rm{t}}^{\rm{2}}}} \right)} - 4{\rm{ft}}} \right\} \end{array} \right] = 0\\ \left[ \begin{array}{l} \left( {12\;{\rm{lb}}} \right)y\\ - \left( {{\rm{30}}\;{\rm{lb}}} \right)\left\{ {\sqrt {{y^2} + \left( {16\;{\rm{f}}{{\rm{t}}^{\rm{2}}}} \right)} - 4{\rm{ft}}} \right\} \end{array} \right] = 0\\ \left( {12\;{\rm{lb}}} \right)y = \left( {{\rm{30}}\;{\rm{lb}}} \right)\left\{ {\sqrt {{y^2} + \left( {16\;{\rm{f}}{{\rm{t}}^{\rm{2}}}} \right)} - 4{\rm{ft}}} \right\}\\ 0.4y = \left[ {\sqrt {{y^2} + \left( {16\;{\rm{f}}{{\rm{t}}^{\rm{2}}}} \right)} - 4{\rm{ft}}} \right] \end{array}\]

On further solving the above equation,

\[\begin{array}{c} \left[ {0.4y + 4} \right] = \sqrt {{y^2} + 16} \\ \left[ {0.16{y^2} + 16 + 3.2y} \right] = \left[ {{y^2} + 16} \right]\\ \left[ {0.84{y^2} - 3.2y} \right] = 0\\ \left[ {y\left( {0.84y - 3.2} \right)} \right] = 0 \end{array}\]
 
Step 5

We will solve the above equation and obtain the value of $y$ (maximum distance $A$ will fall before its motion is momentarily stopped).

The distance $y$ cannot be equal to zero. So,

\[\begin{array}{c} \left( {0.84y - 3.2} \right)\;{\rm{ft}} = 0\\ y \approx 3.81\;{\rm{ft}} \end{array}\]