Step 1

We are given the following data:

The mass of ball is $m = 0.5\;{\rm{kg}}$.

The initial velocity of ball is ${v_A} = 10\;{\rm{m/s}}$.

The coefficient of restitution is $e = 0.5$.

The value of angle is $\theta = 30^\circ $.

The horizontal distance between point $A$ and wall is ${s_{\rm{x}}} = 3\;{\rm{m}}$.

The vertical distance of point $A$ from the horizontal is ${\left( {{s_0}} \right)_{\rm{y}}} = 1.5\;{\rm{m}}$.

We are asked to determine:

(a) The velocity at which it strikes the wall at $B$.

(b) The velocity at which it rebounds from the wall.

(c) The distance $s$ from the wall to where it strikes the ground at $C$.

 
Step 2

(a)

The formula to calculate the time before the impact is given by,

\[\begin{array}{c} {s_{\rm{x}}} = {v_{\rm{x}}}t\\ {s_{\rm{x}}} = \left( {{v_A}\cos \theta } \right)\left( t \right) \end{array}\]

Here, $t$ represent the time before the impact.

Substitute all the known values in the above formula.

\[\begin{array}{c} \left( {3\;{\rm{m}}} \right) = \left[ {\left( {10\;{\rm{m/s}}} \right)\left( {\cos 30^\circ } \right)\left( t \right)} \right]\\ \left( {8.66025\;{\rm{m/s}}} \right)t = 3\;{\rm{m}}\\ t = {\rm{0}}{\rm{.3464}}\;{\rm{s}} \end{array}\]
 
Step 3

The formula to calculate the velocity of the ball before impact is given by,

\[{v_{\rm{y}}} = \left[ {{v_A}\sin \theta - gt} \right]\]

Here, $g$ is the gravitational acceleration $g = 9.81\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}$

Substitute all the known values in the above formula.

\[\begin{array}{c} {v_{\rm{y}}} = \left[ {\left( {10\;{\rm{m/s}}} \right)\left( {\sin 30^\circ } \right) - \left( {9.81\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}} \right)\left( {0.3464\;\sec } \right)} \right]\\ {v_{\rm{y}}} = \left[ {5 - 3.39818} \right]\;{\rm{m/s}}\\ {v_{\rm{y}}} = 1.602\;{\rm{m/s}} \end{array}\]
 
Step 4

The formula to calculate the vertical position of the ball above the ground at point $B$ is given by,

\[{s_{\rm{y}}} = {\left( {{s_0}} \right)_{\rm{y}}} + \left( {{v_A}\sin \theta } \right)\left( t \right) - \frac{1}{2}g{t^2}\]

Here, ${s_{\rm{y}}}$ represent the vertical position of the ball above the ground at point $B$.

Substitute all the known values in the above formula.

\[\begin{array}{c} {s_{\rm{y}}} = \left[ \begin{array}{l} \left( {1.5\;{\rm{m}}} \right) + \left( {10\;{\rm{m/s}}} \right)\left( {\sin 30^\circ } \right)\left( {0.3464\;\sec } \right)\\ - \frac{1}{2}\left( {9.81\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}} \right){\left( {0.3464\;{\mathop{\rm s}\nolimits} } \right)^2} \end{array} \right]\\ {s_{\rm{y}}} = 2.643\;{\rm{m}} \end{array}\]
 
Step 5

The formula to calculate the velocity at point $B$ when it strikes the wall is given by,

\[{\left( {{v_{\rm{b}}}} \right)_1} = \sqrt {{{\left( {{v_A}\cos \theta } \right)}^2} + {{\left( {{v_{\rm{y}}}} \right)}^2}} \]

Substitute all the known values in the above formula.

\[\begin{array}{c} {\left( {{v_{\rm{b}}}} \right)_1} = \sqrt {{{\left[ {\left( {10\;{\rm{m/s}}} \right)\left( {\cos 30^\circ } \right)} \right]}^2} + {{\left( {1.602\;{\rm{m/s}}} \right)}^2}} \\ {\left( {{v_{\rm{b}}}} \right)_1} = \sqrt {77.5664\;{{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{s}}^{\rm{2}}}} \\ {\left( {{v_{\rm{b}}}} \right)_1} = 8.807\;{\rm{m/s}} \end{array}\]
 
Step 6

The formula to calculate the angle with the horizontal is given by,

\[\theta ' = {\tan ^{ - 1}}\left( {\frac{{{v_{\rm{y}}}}}{{{v_A}\cos \theta }}} \right)\]

Here, $\theta '$ represent the angle of the velocity at point $B$ when it strikes the wall with the horizontal.

Substitute all the known values in the above formula.

\[\begin{array}{c} \theta ' = {\tan ^{ - 1}}\left( {\frac{{1.602\;{\rm{m/s}}}}{{\left( {10\;{\rm{m/s}}} \right)\left( {\cos 30^\circ } \right)}}} \right)\\ \theta ' = {\tan ^{ - 1}}\left( {0.18498} \right)\\ \theta ' = 10.48^\circ \end{array}\]
 
Step 7

(b)

We will draw the free-body diagram of the ball when strike the wall.

Here, $\phi $ represent the angle with the horizontal of the final velocity of the ball after strike with wall.

We will apply the conservation of linear momentum to obtain the equation of ${\left( {{v_{\rm{b}}}} \right)_2}$.

\[\begin{array}{c} m{\left( {{v_{\rm{b}}}} \right)_{{\rm{1y}}}} = m{\left( {{v_{\rm{b}}}} \right)_{{\rm{2y}}}}\\ {\left( {{v_{\rm{b}}}} \right)_{{\rm{1y}}}} = {\left( {{v_{\rm{b}}}} \right)_{{\rm{2y}}}}\\ {\left( {{v_{\rm{b}}}} \right)_{{\rm{1y}}}} = {\left( {{v_{\rm{b}}}} \right)_{\rm{2}}}\sin \phi \end{array}\]

Here, ${\left( {{v_{\rm{b}}}} \right)_{{\rm{1y}}}}$ represent the vertical component of the initial velocity of the ball and ${\left( {{v_{\rm{b}}}} \right)_{{\rm{2y}}}}$ represent the vertical component of the final velocity of the ball after strike with wall.

Substitute all the known values in the above formula.

\[\left( {1.602\;{\rm{m/s}}} \right) = {\left( {{v_{\rm{b}}}} \right)_{\rm{2}}}\sin \phi \]
 
Step 8

The formula of the coefficient of restitution is given by,

\[\begin{array}{c} e = \frac{{{{\left( {{v_{\rm{w}}}} \right)}_2} - {{\left( {{v_{\rm{b}}}} \right)}_{{\rm{2x}}}}}}{{{{\left( {{v_{\rm{b}}}} \right)}_{{\rm{1x}}}} - {{\left( {{v_{\rm{w}}}} \right)}_1}}}\\ e = \left( {\frac{{{{\left( {{v_{\rm{w}}}} \right)}_2} - \left[ { - {{\left( {{v_{\rm{b}}}} \right)}_2}\cos \phi } \right]}}{{{v_{\rm{A}}}\cos \theta - {{\left( {{v_{\rm{w}}}} \right)}_1}}}} \right) \end{array}\]

Here, ${\left( {{v_{\rm{w}}}} \right)_1}$ and ${\left( {{v_{\rm{w}}}} \right)_2}$ represent the initial and final velocity of the wall which will be equal to zero because wall is stationary and ${\left( {{v_{\rm{b}}}} \right)_{{\rm{1x}}}}$ and ${\left( {{v_{\rm{b}}}} \right)_{{\rm{2x}}}}$ are the initial and final horizontal component of velocity of the ball.

Substitute all the known values in the above formula.

\[\begin{array}{c} 0.5 = \frac{{0 - \left[ { - {{\left( {{v_{\rm{b}}}} \right)}_2}\cos \phi } \right]}}{{\left( {10\;{\rm{m/s}}} \right)\left( {\cos 30^\circ } \right)}}\\ {\left( {{v_{\rm{b}}}} \right)_2}\cos \phi = 4.33\;{\rm{m/s}} \end{array}\]
 
Step 9

Divide the equation $\left( {1.602\;{\rm{m/s}}} \right) = {\left( {{v_{\rm{b}}}} \right)_{\rm{2}}}\sin \phi $ by the equation ${\left( {{v_{\rm{b}}}} \right)_2}\cos \phi = 4.33$ to obtain $\phi $.

\[\begin{array}{c} \frac{{\left( {1.602\;{\rm{m/s}}} \right)}}{{\left( {4.33\;{\rm{m/s}}} \right)}} = \frac{{{{\left( {{v_{\rm{b}}}} \right)}_{\rm{2}}}\sin \phi }}{{{{\left( {{v_{\rm{b}}}} \right)}_2}\cos \phi }}\\ \tan \phi = 0.369\\ \phi = 20.3^\circ \end{array}\]
 
Step 10

Substitute the value $\phi = 20.3^\circ $ in the equation $\left( {1.602\;{\rm{m/s}}} \right) = {\left( {{v_{\rm{b}}}} \right)_{\rm{2}}}\sin \phi $ to obtain ${\left( {{v_{\rm{b}}}} \right)_2}$.

\[\begin{array}{c} \left( {1.602\;{\rm{m/s}}} \right) = {\left( {{v_{\rm{b}}}} \right)_{\rm{2}}}\sin 20.3^\circ \\ {\left( {{v_{\rm{b}}}} \right)_{\rm{2}}} = \frac{{\left( {1.602\;{\rm{m/s}}} \right)}}{{\sin 20.3^\circ }}\\ {\left( {{v_{\rm{b}}}} \right)_{\rm{2}}} = 4.617\;{\rm{m/s}} \end{array}\]
 
Step 11

(c)

The formula to calculate the time when the ball reaches the vertical position after the impact is given by,

\[ - {s_{\rm{y}}} = \left[ {{{\left( {{v_{\rm{b}}}} \right)}_2}\sin \phi } \right]\left( {{t_1}} \right) - \frac{1}{2}g{t_1}^2\]

Here, ${t_1}$ is the time at which the ball reaches the vertical position after the impact.

Substitute all the known values in the above formula.

\[\begin{array}{c} \left( { - 2.643\;{\rm{m}}} \right) = \left[ \begin{array}{l} \left( {4.617\;{\rm{m/s}}} \right)\left( {\sin 20.3^\circ } \right)\left( {{t_1}} \right)\\ - \frac{1}{2}\left( {9.81\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}} \right){\left( {{t_1}} \right)^2} \end{array} \right]\\ 0 = 4.905{t_1}^2 - 1.601{t_1} - 2.643\\ {t_1} = \frac{{ - \left( { - 1.601} \right) \pm \sqrt {{{\left( { - 1.601} \right)}^2} - 4\left( {4.905} \right)\left( { - 2.643} \right)} }}{{2\left( {4.905} \right)}}\\ {t_1} = 0.9153\;{\rm{s,}} - {\rm{0}}{\rm{.5887}}\;{\rm{s}} \end{array}\]

The value of time cannot be negative. So, the time would be ${t_1} = 0.9153\;{\rm{s}}$.

 
Step 12

The formula to calculate the horizontal distance after the impact is given by,

\[{s_{\rm{x}}} = s = {\left( {{v_{\rm{b}}}} \right)_2}\left( {\cos \phi } \right)\left( {{t_1}} \right)\]

Here, ${s_{\rm{x}}}$ represent the horizontal distance after the impact.

Substitute all the known values in the above formula.

\[\begin{array}{c} s = \left( {4.617\;{\rm{m/s}}} \right)\left( {\cos 20.3^\circ } \right)\left( {0.9153\;{\rm{s}}} \right)\\ s = 3.96\;{\rm{m}} \end{array}\]