Step 1

We are given a baseball player $A$ hitting the baseball at an angle from the horizontal. When the ball is directly overhead of player $B$ he begins to run under it.

The initial velocity of the ball at $A$ is ${v_A} = 40{\rm{ ft/s}}$.

The angle of the baseball from the horizontal at point $A$ is ${\theta _A} = 60^\circ $.

We are asked to determine the constant speed at which player $B$ must run and the distance $d$ in order to make the catch at the same elevation at which the ball was hit.

 
Step 2

The diagram showing the path of the baseball is given below:

Here, $d$ is the distance travels by the player in order to make the catch at the same elevation at which the ball was hit.

 
Step 3

The horizontal component of initial velocity of baseball is given by,

\[\begin{array}{c} {\left( {{v_A}} \right)_x} = {v_A}\cos {\theta _A}\\ {\left( {{v_A}} \right)_x} = 40{\rm{ ft/s}}\left( {\cos 60^\circ } \right)\\ {\left( {{v_A}} \right)_x} = 20{\rm{ ft/s}} \end{array}\]

The vertical component of initial velocity of baseball is given by,

\[\begin{array}{c} {\left( {{v_A}} \right)_y} = {v_A}\sin {\theta _A}\\ {\left( {{v_A}} \right)_y} = 40{\rm{ ft/s}}\left( {\sin 60^\circ } \right)\\ {\left( {{v_A}} \right)_y} = 34.64{\rm{ ft/s}} \end{array}\]
 
Step 4

Using the second equation of motion, the vertical displacement of the baseball at point $A$ is given by,

\[{\left( {{s_A}} \right)_y} = {\left( {{v_A}} \right)_y}t - \frac{1}{2}g{t^2}\]

Here, ${\left( {{s_A}} \right)_y}$ is the final displacement of baseball at point $A$, its value is ${\left( {{s_A}} \right)_y} = 0$, $t$ is the time of flight of baseball, and $g$ is the acceleration due gravity, its value is $g = 32.2{\rm{ ft/}}{{\rm{s}}^2}$.

Substituting the values in the above expression, we get,

\[\begin{array}{c} 0 = 34.64t - \frac{1}{2}\left( {32.2} \right){t^2}\\ 16.1{t^2} = 34.64t = 0\\ t = \left( {\frac{{34.64}}{{16.1}}} \right)\\ t = 2.152{\rm{ s}} \end{array}\]
 
Step 5

Using the relation between distance and velocity, the total displacement between point $A$ and $C$ is given by,

\[\begin{array}{c} x + d = {\left( {{v_A}} \right)_x}t\\ \left( {15{\rm{ ft}}} \right) + d = \left( {20{\rm{ ft/s}}} \right) \times \left( {{\rm{2}}{\rm{.152 s}}} \right)\\ \left( {15{\rm{ ft}}} \right) + d = \left( {43.04{\rm{ ft}}} \right)\\ d = 28.04{\rm{ ft}} \end{array}\]

Since, the player $B$ runs at a same speed as the horizontal component of velocity of the baseball.

So, the velocity of player $B$ is given by,

\[\begin{array}{c} {v_B} = {\left( {{v_A}} \right)_x}\\ {v_B} = \left( {40{\rm{ ft/s}}} \right)\left( {\cos 60^\circ } \right)\\ {v_B} = 20{\rm{ ft/s}} \end{array}\]