Chapter 4, Problem 27P

(a)

To determine

The speed of the ball at time $t=3.2\text{s}$.

Answer to Problem 27P

The speed of the ball at time $t=3.2\text{s}$ is $\underset{\_}{37\text{m}/\text{s}}$.

Explanation of Solution

Write the expression for horizontal component of speed.

    $v_x=v_0\cos \theta$        (I)

Here, $v_x$ is the horizontal component of speed, $v_0$ is the initial speed, and $\theta$ is the angle relative to the ground.

Write the expression for vertical component of speed.

    $v_y=v_0\sin \theta -gt$        (II)

Here, $v_x$ is the horizontal component of speed, $g$ is the acceleration due to gravity, and $t$ is the time.

Write the expression for magnitude of speed.

    $v=\sqrt{v_x^2+v_y^2}$        (III)

Here, $v$ is the magnitude of speed.

Conclusion:

Substitute $45\text{m}/\text{s}$ for $v_0$, and $35°$ for $\theta$ in equation (I), to find $v_x$.

    $\begin{array}{c}{v}_x=\left(45\text{m}/\text{s}\right)\cos 35°\\ =36.9\text{m}/\text{s}\end{array}$

Substitute $45\text{m}/\text{s}$ for $v_0$, $9.8\text{m}/{\text{s}}^2$ for $g$, $3.2\text{s}$ for $t$, and $35°$ for $\theta$ in equation (II), to find $v_y$.

    $\begin{array}{c}{v}_y=\left(45\text{m}/\text{s}\right)\sin 35°-\left(9.8\text{m}/{\text{s}}^2\right)\left(3.2\text{s}\right)\\ =-5.55\text{m}/\text{s}\end{array}$

Substitute $-5.55\text{m}/\text{s}$ for $v_y$, and $36.9\text{m}/\text{s}$ for $v_x$ in equation (III), to find $v$.

    $\begin{array}{c}v=\sqrt{{\left(36.9\text{m}/\text{s}\right)}^2+{\left(-5.55\text{m}/\text{s}\right)}^2}\\ =37\text{m}/\text{s}\end{array}$

Therefore, the speed of the ball at time $t=3.2\text{s}$ is $\underset{\_}{37\text{m}/\text{s}}$.

(b)

To determine

The angle $\theta$, velocity $\overrightarrow{v}$ makes with the $x$ axis.

Answer to Problem 27P

The angle $\theta$, the velocity $\overrightarrow{v}$ makes with the $x$ axis is $\underset{\_}{-8.6°}$.

Explanation of Solution

Write the expression for angle $\theta$, the velocity $\overrightarrow{v}$ makes with the $x$ axis.

    $\tan \theta =\frac{v_y}{v_x}$

Rearrange the above equation.

  $\theta ={\tan }^{-1}\left(\frac{v_y}{v_x}\right)$        (IV)

Conclusion:

Substitute $-5.55\text{m}/\text{s}$ for $v_y$, and $36.9\text{m}/\text{s}$ for $v_x$ in equation (IV), to find $\theta$.

    $\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{-5.55\text{m}/\text{s}}{36.9\text{m}/\text{s}}\right)\\ =-8.6°\end{array}$

Therefore, the angle $\theta$, the velocity $\overrightarrow{v}$ makes with the $x$ axis is $\underset{\_}{-8.6°}$.

(c)

To determine

The value of time $t$ for which the speed is minimum.

Answer to Problem 27P

The value of time $t$ for which the speed is minimum is $\underset{\_}{2.6\text{s}}$.

Explanation of Solution

Write the expression for vertical component of speed.

    $v_y=v_0\sin \theta -gt$

Here, $v_x$ is the horizontal component of speed, $g$ is the acceleration due to gravity, and $t$ is the time.

Rearrange the above equation, to find $t$.

    $\begin{array}{c}{v}_0\sin \theta -v_y=gt\\ t=\frac{v_0\sin \theta -v_y}{g}\end{array}$        (V)

Conclusion:

Substitute $45\text{m}/\text{s}$ for $v_0$, $9.8\text{m}/{\text{s}}^2$ for $g$, $0\text{m}/\text{s}$ for $v_y$, and $35°$ for $\theta$ in equation (V), to find $t$.

    $\begin{array}{c}t=\frac{\left(45\text{m}/\text{s}\right)\sin 35°-\left(0\text{m}/\text{s}\right)}{\left(9.8\text{m}/\text{s}\right)}\\ =2.6\text{s}\end{array}$

Therefore, The value of time $t$ for which the speed is minimum is $\underset{\_}{2.6\text{s}}$.