Chapter 6, Problem 53A

To determine

The speed given by the player to ball to make the shot.

Answer to Problem 53A

The speed given by the player to the ball in order to make the shot is $11.\text{8 m/s}$ .

Explanation of Solution

Given:

The angle made by the ball with the horizontal is $\theta =51°$ .

The ball covers a horizontal distance of $14\text{m}$ from the basket.

Formula used:

The relation between the velocity and time of a moving object is

  $u=\frac{d}{t}$

Here, $u$ is the velocity of the moving object, $d$ is the distance covered and $t$ is the time taken by the object.

The equation of motion is

  $y=ut+\frac{1}{2}a{t}^2$

Here, $u$ is the velocity of the moving object, $y$ is the vertical distance covered by the object, $t$ is the time taken by the object, and $a$ is the acceleration of the moving object.

Calculation:

Consider the basket ball player hits the ball with the initial velocity $v_{\text{i}}$ at an angle $\theta =51°$ follows the parabolic path and reaches the target.

The horizontal component $\left(u_{\text{xi}}\right)$ and the vertical component $\left(u_{\text{yi}}\right)$ of the initial velocity are,

  $u_{\text{xi}}=v_{\text{i}}\cos \theta$

  $u_{\text{yi}}=v_{\text{i}}\sin \theta$

The time $\left(t\right)$ taken by the ball to reach the target is

  $\begin{array}{c}x=\text{ (}{u}_{\text{x}}\text{)}t\\ x=\text{(}{u}_{\text{i}}\text{ cos}\theta \text{)}t\\ t=\frac{x}{\text{(}{u}_{\text{i}}\text{ cos}\theta \text{)}}\end{array}$

To find the initial velocity of the ball, apply the kinematic equation,

  $\begin{array}{c}y=u_{\text{y}}t-\frac{1}{2}g{t}^{\text{2}}\\ y=(v_{\text{i}}\sin \theta )t-\frac{1}{2}g{t}^{\text{2}}\end{array}$

At the target point, the ball reaches the ground surface. Then, the vertical distance $\left(y\right)$ covered by the ball at the target point is zero.

  $\begin{array}{l}0=(v_{\text{i}}\sin \theta )t-\frac{1}{2}g{t}^{\text{2}}\\ (v_{\text{i}}\sin \theta )t-\frac{1}{2}g{t}^{\text{2}}=0\end{array}$

Substitute $t=\frac{x}{(v_{\text{i}}\cos \theta )}$ in the above equation.

  $\begin{array}{l}(v_{\text{i}}\sin \theta )\frac{x}{(v_{\text{i}}\cos \theta )}-\frac{1}{2}g{\left(\frac{x}{(v_{\text{i}}\cos \theta )}\right)}^{\text{2}}=0\\ (\tan \theta )x-\frac{1}{2}g\left[\frac{x^2}{v_{\text{i}}^2{\cos }^2\theta }\right]=0\\ (\tan \theta )x-\frac{1}{2}g\left[\frac{x^2}{v_{\text{i}}^2{\cos }^2\theta }\right]=0\\ (\tan \theta )x=\frac{1}{2}g\left(\frac{x^2}{v_{\text{i}}^2{\cos }^2\theta }\right)\\ (\tan \theta )(v_{\text{i}}^2{\cos }^2\theta )=\frac{xg}{2}\end{array}$

  $\begin{array}{l}(v_{\text{i}}\sin \theta )\frac{x}{(v_{\text{i}}\cos \theta )}-\frac{1}{2}g{\left(\frac{x}{(v_{\text{i}}\cos \theta )}\right)}^{\text{2}}=0\\ (\tan \theta )x-\frac{1}{2}g\left(\frac{x^2}{v_{\text{i}}^2{\cos }^2\theta }\right)=0\\ (\tan \theta )x=\frac{1}{2}g\left(\frac{x^2}{v_{\text{i}}^2{\cos }^2\theta }\right)\\ (\tan \theta )(v_{\text{i}}^2{\cos }^2\theta )=\frac{xg}{2}\end{array}$

  $\begin{array}{l}\frac{\sin \theta }{\cos \theta }\left(v_{\text{i}}{}^{\text{2}}{\cos }^2\theta \right)=\frac{x}{2}g\\ v_{\text{i}}{}^2\sin \theta \cos \theta =\frac{x}{2}g\\ v_{\text{i}}{}^2=\frac{x}{2}\times \left(\frac{g}{\sin \theta \cos \theta }\right)\\ v_{\text{i}}=\sqrt{\frac{gx}{2\sin \theta \cos \theta }}\end{array}$

Take the acceleration due to gravity as $g=\text{9}{\text{.8m/s}}^{\text{2}}$ .

Substitute the values of distance as $x=14.0\text{m}$ , angle of projection as $\theta =51.0°$ , and acceleration due to gravity as $g=\text{9}{\text{.8m/s}}^{\text{2}}$ .

  $\begin{array}{c}v=\sqrt{\frac{(14.0)(\text{9}{\text{.8 m/s}}^{\text{2}})}{2\sin ({51.0}^{\circ })\cos ({51.0}^{\circ })}}\\ =11.{\text{8 m/s}}^{\text{2}}\end{array}$

Conclusion:

Therefore, the speed given by the player in order to make the shot is $11.\text{8 m/s}$ .