Step 1

We are given that the weight of brick is $W = 2\;{\rm{lb}}$, the velocity of brick at point A is ${v_A} = 5\;{\rm{ft/s}}$.

We are asked to calculate the speed of brick at point B, the horizontal distance from wall where the brick falls and the speed of brick at point C.

We have the height of brick at point A is ${h_A} = \left( {30\;{\rm{ft}} + 15\;{\rm{ft}}} \right) = 45\;{\rm{ft}}$.

We have the height of brick at point B is ${h_B} = 30\;{\rm{ft}}$.

 
Step 2

To calculate the work done by gravitational force to move from point A to B we use the formula:

\[U = W\left( {{h_A} - {h_B}} \right)\]
 
Step 3

Substitute the known values in the formula:

\[\begin{array}{c} U = \left( {{\rm{2}}\;{\rm{lb}}} \right)\left( {45\;{\rm{ft}} - 30\;{\rm{ft}}} \right)\\ = \left( {{\rm{2}}\;{\rm{lb}}} \right)\left( {{\rm{15}}\;{\rm{ft}}} \right)\\ = 30\;{\rm{lb}} \cdot {\rm{ft}} \end{array}\]
 
Step 4

According to work energy principle,

\[{T_A} + U = {T_B}\] …… (1)
 
Step 5

To calculate the kinetic energy of the brick at point A we use the formula:

\[{T_A} = \frac{1}{2}mv_A^2\]
 
Step 6

To calculate the kinetic energy of the brick at point B we use the formula:

\[{T_B} = \frac{1}{2}mv_B^2\]
 
Step 7

Substitute the known values in equation (1):

\[\frac{1}{2}mv_A^2 + U = \frac{1}{2}mv_B^2\]…… (2)

 
Step 8

To calculate the mass of brick we use the formula:

\[m = \frac{W}{g}\]
 
Step 9

Substitute the known values in equation (2):

\[\frac{1}{2}\left( {\frac{W}{g}} \right)v_A^2 + U = \frac{1}{2}\left( {\frac{W}{g}} \right)v_B^2\]
 
Step 10

Substitute the known values in equation:

\[\begin{array}{c} \frac{1}{2}\frac{{\left( {{\rm{2}}\;{\rm{lb}}} \right)}}{{\left( {32.2\;{\rm{ft/}}{{\rm{s}}^{\rm{2}}}} \right)}}{\left( {{\rm{5}}\;{\rm{ft/s}}} \right)^2} + 30\;{\rm{lb}} \cdot {\rm{ft}} = \frac{1}{2}\frac{{\left( {{\rm{2}}\;{\rm{lb}}} \right)}}{{\left( {32.2\;{\rm{ft/}}{{\rm{s}}^2}} \right)}}v_B^2\\ 0.7764\;{\rm{lb}} \cdot {\rm{ft}} + 30\;{\rm{lb}} \cdot {\rm{ft}} = \left( {0.0310\;{\rm{lb}} \cdot {{\rm{s}}^{\rm{2}}}{\rm{/ft}}} \right)v_B^2\\ v_B^2 = \frac{{\left( {30.7764\;{\rm{lb}} \cdot {\rm{ft}}} \right)}}{{\left( {0.0310\;{\rm{lb}} \cdot {{\rm{s}}^{\rm{2}}}{\rm{/ft}}} \right)}}\\ {v_B} = 31.5\;{\rm{ft/s}} \end{array}\]
 
Step 11

To calculate the horizontal distance covered by brick we use the formula:

\[d = {v_x}t\]…… (3)
 
Step 12

To calculate the horizontal component of velocity of brick we use the formula:

\[{v_x} = {v_B}\cos \theta \]
 
Step 13

Substitute the known value in equation (3):

\[d = \left( {{v_B}\cos \theta } \right)t\]…… (4)
 
Step 14

From the given diagram,

\[\cos \theta = \frac{4}{5}\]
 
Step 15

Substitute the known value in equation (4):

\[\begin{array}{c} d = \left( {31.5\;{\rm{ft/s}}} \right)\left( {\frac{4}{5}} \right)t\\ = \left( {25.2\;{\rm{ft/s}}} \right)t \end{array}\]…… (5)
 
Step 16

To calculate the vertical distance covered by brick we use the formula:

\[{h_B} = {v_y}t + \frac{1}{2}g{t^2}\]…… (6)
 
Step 17

To calculate the vertical component of velocity of brick we use the formula:

\[{v_y} = {v_B}\sin \theta \]
 
Step 18

Substitute the known value in equation (6):

\[{h_B} = \left( {{v_B}\sin \theta } \right)t + \frac{1}{2}g{t^2}\] …… (7)
 
Step 19

From the given diagram,

\[\sin \theta = \frac{3}{5}\]
 
Step 20

Substitute the known value in equation (7):

\[\begin{array}{c} 30\;{\rm{ft}} = \left( {{\rm{31}}{\rm{.5}}\;{\rm{ft/s}}} \right)\left( {\frac{3}{5}} \right)t + \frac{1}{2}\left( {{\rm{32}}{\rm{.2}}\;{\rm{ft/}}{{\rm{s}}^{\rm{2}}}} \right){t^2}\\ \left( {16.1\;{\rm{ft/}}{{\rm{s}}^2}} \right){t^2} + \left( {18.9\;{\rm{ft/s}}} \right)t - 30\;{\rm{ft}} = 0\\ t = \frac{{ - \left( {18.9\;{\rm{ft/s}}} \right) + \sqrt {{{\left( {{\rm{18}}{\rm{.9}}\;{\rm{ft/s}}} \right)}^2} - 4\left( {16.1\;{\rm{ft/}}{{\rm{s}}^2}} \right)\left( { - 30\;{\rm{ft}}} \right)} }}{{2\left( {16.1\;{\rm{ft/}}{{\rm{s}}^2}} \right)}}\\ = 0.898\;{\rm{s}} \end{array}\]
 
Step 21

Substitute the known value in equation (5):

\[\begin{array}{c} d = \left( {25.2\;{\rm{ft/s}}} \right)\left( {0.898\;{\rm{s}}} \right)\\ = 22.6\;{\rm{ft}} \end{array}\]
 
Step 22

According to work energy principle,

\[{T_A} + {U_A} = {T_C}\] …… (8)
 
Step 23

To calculate the potential energy of the brick at point A we use the formula:

\[{U_A} = W{h_A}\]
 
Step 24

To calculate the kinetic energy of the brick at point C we use the formula:

\[{T_C} = \frac{1}{2}mv_C^2\]
 
Step 25

Substitute the known values in equation (1):

\[\begin{array}{c} \frac{1}{2}\left( {\frac{W}{g}} \right)v_A^2 + W{h_A} = \frac{1}{2}\left( {\frac{W}{g}} \right)v_C^2\\ v_C^2 = v_A^2 + 2g{h_A}\\ {v_C} = \sqrt {v_A^2 + 2g{h_A}} \end{array}\]
 
Step 26

Substitute the known values in the equation:

\[\begin{array}{c} {v_C} = \sqrt {{{\left( {{\rm{5}}\;{\rm{ft/s}}} \right)}^2} + 2\left( {{\rm{32}}{\rm{.2}}\;{\rm{ft/}}{{\rm{s}}^{\rm{2}}}} \right)\left( {45\;{\rm{ft}}} \right)} \\ = \sqrt {25\;{\rm{f}}{{\rm{t}}^{\rm{2}}}{\rm{/}}{{\rm{s}}^2} + 2898\;{\rm{f}}{{\rm{t}}^{\rm{2}}}{\rm{/}}{{\rm{s}}^2}} \\ = \sqrt {2923\;{\rm{f}}{{\rm{t}}^{\rm{2}}}{\rm{/}}{{\rm{s}}^{\rm{2}}}} \\ = 54.1\;{\rm{ft/s}} \end{array}\]