Chapter 2, Problem 31QAP

To determine

(a)

The acceleration of a thrown base-ball.

Answer to Problem 31QAP

The acceleration of the thrown baseball = $\frac{{\text{V}}_f^2}{2d}$

where $V_f$= the velocity of the baseball at the time of release by the pitcher

d = the displacement of the ball from behind the body of the pitcher to the point where it is released (Refer the picture below)

Explanation of Solution

Given:

During this calculation it is assumed that the baseball undergoes constant linear acceleration. Also, it is assumed that the linear velocity of the baseball when it is thrown is $V_f$(ms-1) and that the displacement of the baseball during the throwing process by the pitcher is d (in meters).

Formula used:

  $\begin{array}{l}{V}_x^2{\text{ = V}}_{x_0}^2{\text{ + 2a}}_x{\text{(x-x}}_0)\\ V_x\text{ = velocity at position x of an object in linear motion with constant accleration}\\ V_{x_0}{\text{ =velocity at position x}}_0\text{ of an object in linear motion with constant accleration}\\ {\text{a}}_x\text{ = constant accleration of the object}\end{array}$

Calculation:

During the throwing process the baseball is at rest in pitcher's hand. This means that that the initial velocity of the baseball is 0 ms-1. Assume that the pitcher releases the baseball at a linear velocity of $V_f$ms-1.

Also, the difference between the two positions is equal to d. Hence substituting to the above equation;

  $\begin{array}{l}{V}_x=V_f(m{s}^{-1}\text{) }{V}_{x_0}={\text{ 0 ms}}^{-1}{\text{ (x-x}}_0)=d(m)\\ {\text{V}}_f^2{\text{ (ms}}^{-1}{\text{)}}^2{\text{ = (0 ms}}^{-1}{)}^2+2a_{x}d(m)\\ \Rightarrow 2a_{x}d(m)={\text{V}}_f^2{(}^m\\ \Rightarrow a_x=\frac{{\text{V}}_f^2}{2d}{\text{ ms}}^{-2}\end{array}$

Conclusion:

The acceleration of the thrown baseball = $\frac{{\text{V}}_f^2}{2d}$(ms-2)

]

Where $V_f$= the velocity of the baseball at the time of released by the pitcher

d = the displacement of the ball from behind the body of the pitcher to the point where it is released

To determine

(b)

The acceleration of a kicked soccer ball.

Answer to Problem 31QAP

The acceleration of the thrown baseball = - ( $\frac{{\text{V}}_f^2}{2d}$)

where $V_f$= the velocity of the baseball at the time of release by footballer

d = the displacement of the soccer ball

Explanation of Solution

Given:

During this calculation it is assumed that the soccer ball undergoes constant linear acceleration. Also it is assumed that the linear velocity of the baseball when it is thrown is $V_f$(ms-1) and that the displacement of the soccer ball is d (in meters).

Formula used:

  $\begin{array}{l}{V}_x^2{\text{ = V}}_{x_0}^2{\text{ + 2a}}_x{\text{(x-x}}_0)\\ V_x\text{ = velocity at position x of an object in linear motion with constant accleration}\\ V_{x_0}{\text{ =velocity at position x}}_0\text{ of an object in linear motion with constant accleration}\\ {\text{a}}_x\text{ = constant accleration of the object}\end{array}$

Calculation:

It is assumed that player A kicks the soccer ball with a liner velocity and the player B catches the soccer ball. When player B catches the soccer ball the velocity of the ball would be zero. It is assumed that the ball travels a distance of d during this process.

  $\begin{array}{l}{V}_x=0(m{s}^{-1}\text{) }{V}_{x_0}=V_f{\text{ ms}}^{-1}{\text{ (x-x}}_0)=d(m)\\ {\text{0}}^2{\text{ (ms}}^{-1}{\text{)}}^2\text{ =(}{V}_f{\text{)}}^2{\text{ ( ms}}^{-1}{)}^2+2a_{x}d(m)\\ \Rightarrow 2a_{x}d(m)={\text{ - V}}_f^2{(}^m\\ \Rightarrow a_x=-(\frac{{\text{V}}_f^2}{2d}{\text{) ms}}^{-2}\end{array}$

Note in this case the final velocity is zero hence the sign of the acceleration is negative.

Conclusion:

The acceleration of the thrown baseball = - ( $\frac{{\text{V}}_f^2}{2d}$)

where $V_f$= the velocity of the baseball at the time of release by footballer

d = the displacement of the soccer ball