Chapter 3, Problem 64P

(a)

To determine

The total time of flight of the ball as observed by the juggler in the train.

Answer to Problem 64P

The total time of flight of the ball as observed by the juggler in the train is found to be $1.00\text{ s}$ .

Explanation of Solution

Given:

The initial velocity of the ball relative to the train

  ${\overrightarrow{v}}_0=4.90\text{ m/s}\widehat{j}$

Formula used:

To determine the time of flight $t$ , the following equation of motion is used.

  $y=v_{0}t+\frac{1}{2}a{t}^2$...........(1)

Here, $y$ is the total displacement in time $t$ and $a$ is the acceleration of the ball.

Calculation:

The ball thrown in a train and the juggler are at rest with respect to the train. Choose a one dimensional coordinate system with the origin on the train and the positive y axis directed upwards. As the ball moves up and returns to the juggler’s hands, its displacement $\Delta y$ along the y direction is zero. The ball moves under the effect of the Earth’s gravitational force, hence its acceleration is equal to the acceleration of free fall, directed downwards along the −y axis.

Therefore,

  $\Delta y=0;a=g=9.81{\text{ m/s}}^2\left(-\widehat{j}\right)$

Substitute these values in equation (1) and calculate the time of flight.

  $\begin{array}{c}\Delta y=v_{0}t+\frac{1}{2}a{t}^2\\ 0=\left(4.90\text{ m/s}\widehat{j}\right)t+\frac{1}{2}\left(-9.81{\text{ m/s}}^2\widehat{j}\right)t^2\\ t=\frac{2\left(4.90\text{ m/s}\widehat{j}\right)}{\left(9.81{\text{ m/s}}^2\widehat{j}\right)}\\ =1.00\text{ s}\end{array}$

Conclusion:

Thus, the total time of flight of the ball as observed by the juggler in the train is found to be $1.00\text{ s}$ .

(b)

To determine

The displacement of the ball during its rise as observed by the juggler.

Answer to Problem 64P

The displacement of the ball during its rise as observed by the juggler, is found to be $\text{1}\text{.22 m}\widehat{j}$ .

Explanation of Solution

Given:

The initial velocity of the ball relative to the train ${\overrightarrow{v}}_0=4.90\text{ m/s}\widehat{j}$

Formula used:

To determine the displacement of the ball , the following equation of motion may be used.

  $v^2=v_0^2+2a\Delta y$............(2)

Calculation:

As the ball moves upwards, it slows down due to the action of the Earth’s gravitational force. At the top most point of its trajectory, is instantaneous velocity v becomes zero.

Substitute the values of variables in the equation (2) and solve for $\Delta y$ .

  $\begin{array}{c}{v}^2=v_0^2+2a\Delta y\\ 0={\left(4.90\text{ m/s}\widehat{j}\right)}^2+2\left(-9.81{\text{ m/s}}^2\widehat{j}\right)\Delta y\\ \Delta y=\frac{{\left(4.90\text{ m/s}\widehat{j}\right)}^2}{2\left(9.81{\text{ m/s}}^2\widehat{j}\right)}\\ =1.22\text{ m}\widehat{j}\end{array}$

Conclusion:

Thus, the displacement of the ball during its rise as observed by the juggler, is found to be $\text{1}\text{.22 m}\widehat{j}$ .

(c)

To determine

The ball’s initial speed as observed by the friend on the ground.

Answer to Problem 64P

The ball’s initial speed as observed by the friend on the ground is found to be $20.6\text{ m/s}$ .

Explanation of Solution

Given:

The initial velocity of the ball relative to the train${\overrightarrow{v}}_0=4.90\text{ m/s}\widehat{j}$

The velocity of the train relative to the ground${\overrightarrow{v}}_t=20.0\text{ m/s}\widehat{i}$

Formula used:

Using a coordinate system with the origin at the ground and the positive x axis along East, a vector diagram is constructed.

  

Figure 1

The person on the ground observes the ball to have a velocity ${\overrightarrow{v}}_{0g}$ , whose magnitude can be determined using Pythagoras theorem.

  $\left|{\overrightarrow{v}}_{0g}\right|=\sqrt{v_0^2+v_t^2}$...........(3)

Calculation:

Substitute the values of variables in equation (3) and calculate the speed of the ball as observed by the person on the ground.

  $\begin{array}{c}\left|{\overrightarrow{v}}_{0g}\right|=\sqrt{v_0^2+v_t^2}\\ =\sqrt{{\left(4.90\text{ m/s}\right)}^2+{\left(20.0\text{ m/s}\right)}^2}\\ =20.6\text{ m/s}\end{array}$

Conclusion:

Thus, the ball’s initial speed as observed by the friend on the ground is found to be $20.6\text{ m/s}$ .

(d)

To determine

The angle of launch of the ball as observed by the person on the ground.

Answer to Problem 64P

The angle of launch of the ball as observed by the person on the ground $13.8°$to East.

Explanation of Solution

Given:

The initial velocity of the ball relative to the train${\overrightarrow{v}}_0=4.90\text{ m/s}\widehat{j}$

The velocity of the train relative to the ground${\overrightarrow{v}}_t=20.0\text{ m/s}\widehat{i}$

Formula used:

Use Figure 1 to calculate the angle $\theta$ made by the ball to horizontal as follows:

  $\theta ={\tan }^{-1}\left(\frac{v_0}{v_t}\right)$...........(4)

Calculation:

Substitute the values of the variables in equation (4) and calculate the angle of launch of the ball as observed by the person on the ground.

  $\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{v_0}{v_t}\right)\\ ={\tan }^{-1}\left(\frac{4.90\text{ m/s}}{20.0\text{ m/s}}\right)\\ =13.8°\end{array}$

Conclusion:

Thus, the angle of launch of the ball as observed by the person on the ground $13.8°$to East.

(e)

To determine

The displacement of the ball during its rise as observed by the person on the ground.

Answer to Problem 64P

The displacement of the ball during its rise as observed by the person on the ground is found to be $\overrightarrow{r}=\left(9.99\widehat{i}+1.22\widehat{j}\right)\text{m}$ .

Explanation of Solution

Given:

The initial velocity of the ball relative to the train${\overrightarrow{v}}_0=4.90\text{ m/s}\widehat{j}$

The velocity of the train relative to the ground${\overrightarrow{v}}_t=20.0\text{ m/s}\widehat{i}$

Formula used:

The displacement of the ball as seen by the person on the ground is given by the expression,

  $\overrightarrow{r}=\Delta x\widehat{i}+\Delta y\widehat{j}$...........(5)

Here, $\Delta x$ and $\Delta y$ are the displacements along the x and y directions.

From Figure 1, it can be seen that the initial velocity ${\overrightarrow{v}}_{0g}$ of the ball as observed by the person has a component ${\overrightarrow{v}}_t$ along the x direction and ${\overrightarrow{v}}_0$ along the y direction. Therefore the velocity ${\overrightarrow{v}}_{0g}$ can be written as

  ${\overrightarrow{v}}_{0g}=v_t\widehat{i}+v_0\widehat{j}$...........(6)

The time $t$ taken by the ball to reach the maximum height is determined using the equation of motion,

  $v=v_0+at$...........(7)

The horizontal component of the ball’s velocity remains constant, since no force acts along the horizontal direction. While, since the acceleration of free fall acts downwards, the vertical component of the ball’s velocity varies with time.

The values of $\Delta x$ and $\Delta y$ are determined using the following expressions:

  $\Delta x=v_{t}t$...........(8)

  $\Delta y=v_{0}t+\frac{1}{2}a{t}^2$...........(9)

Calculation:

The trajectory of the ball as seen by the person on the ground is shown in the diagram below;

  

At the top most point of its trajectory, the vertical component of the ball’s velocity becomes equal to zero. Use equation (7) and calculate the time taken by the ball to reach the top most point of its trajectory.

  $\begin{array}{c}t=\frac{\left(4.90\text{ m/s}\right)}{\left(9.81{\text{ m/s}}^2\right)}\\ =0.499\text{ s}\end{array}$

Calculate the value of $\Delta x$ by substituting the values of variables in equation (8).

  $\begin{array}{c}\Delta x=v_{t}t\\ =\left(20.0\text{ m/s}\right)\left(0.499\text{ s}\right)\\ =9.99\text{ m}\end{array}$

Calculate the value of $\Delta y$ by substituting the values of variables in equation (9).

  $\begin{array}{c}\Delta y=v_{0}t+\frac{1}{2}a{t}^2\\ =\left(4.90\text{ m/s}\right)\left(0.499\text{ s}\right)-\frac{1}{2}\left(9.81{\text{ m/s}}^2\right){\left(0.499\text{ s}\right)}^2\\ =1.22\text{ m}\end{array}$

Substitute the values of $\Delta x$ and $\Delta y$ in equation (5).

  $\begin{array}{c}\overrightarrow{r}=\Delta x\widehat{i}+\Delta y\widehat{j}\\ \overrightarrow{r}=\left(9.99\widehat{i}+1.22\widehat{j}\right)\text{m}\end{array}$

Conclusion:

Thus, the displacement of the ball during its rise as observed by the person on the ground is found to be $\overrightarrow{r}=\left(9.99\widehat{i}+1.22\widehat{j}\right)\text{m}$ .