Chapter 4, Problem 47P

(a)

To determine

Sketch the graphs of x,y, $v_x$ and $v_y$ as functions of time.

Answer to Problem 47P

Figure 1, 2, 3, 4 shows the graphs of x,y, $v_x$ and $v_y$ as functions of time.

Explanation of Solution

Figure 1

Conclusion:

Therefore, Figure 1, 2, 3, 4 shows the graphs of x,y, $v_x$ and $v_y$ as functions of time.

(b)

To determine

The initial velocity of the stone.

Answer to Problem 47P

The initial velocity of the stone is $\underset{\_}{27.6\text{m}/\text{s}}$ at $25°$ above the horizontal.

Explanation of Solution

Write the expression for the initial horizontal component of velocity of the stone.

    $v_{ix}=v_i\cos 25°$        (I)

Here, $v_{ix}$ is the initial horizontal component of velocity of the stone, $v_i$ is the initial velocity.

Write the expression for the initial vertical component of velocity of the stone.

    $v_{iy}=v_i\sin 25°$        (II)

Write the expression for the equation of motion.

    $x=v_{x}t+\frac{1}{2}{a}_x{t}^2$        (III)

Conclusion:

Substitute $105\text{m}$ for $x$, $0$ for $v_x$, $4.2\text{s}$ for $t$  in equation (III) to find $x$.

    $\begin{array}{c}105\text{m}=v_{ix}\left(4.2\text{s}\right)+\frac{1}{2}\left(0\text{m}/{\text{s}}^2\right){\left(4.2\text{s}\right)}^2\\ v_{ix}=\frac{105\text{m}}{4.2\text{s}}\\ =25\text{m}/\text{s}\end{array}$

Substitute $25\text{m}/\text{s}$ for $v_{ix}$ in equation (I) to find $v_i$.

    $\begin{array}{c}{v}_i=\frac{v_{ix}}{\cos 25°}\\ =\frac{25\text{m}/\text{s}}{\cos 25°}\\ =27.6\text{m}\end{array}$

Therefore, the initial velocity of the stone is $\underset{\_}{27.6\text{m}/\text{s}}$ at $25°$ above the horizontal.

(c)

To determine

The initial height $h$.

Answer to Problem 47P

The initial height $h$ is $\underset{\_}{37.5\text{m}}$

Explanation of Solution

Let the initial height of the stone is $h$. Then at the end of the journey the vertical displacement of the stone is $y=-h$.

The vertical component of initial velocity is ,

    $\begin{array}{c}{v}_{iy}=\left(27.6\text{m}/\text{s}\right)\sin 25°\\ =11.664\text{m}/\text{s}\end{array}$

Write the expression for the equation of motion.

    $y=v_{oy}t+\frac{1}{2}{a}_y{t}^2$        (IV)

Conclusion:

At the end of the journey the vertical displacement of the stone is $y=-h$.substitute $11.664\text{m}/\text{s}$ for $v_{iy}$ , $4.2\text{s}$ for $t$, $-9.8\text{m}/{\text{s}}^2$ for $a_y$ in equation (IV) to find $h$.

    $\begin{array}{c}-h=\left(11.664\text{m}/\text{s}\right)\left(4.2\text{s}\right)+\frac{1}{2}\left(-9.8\text{m}/{\text{s}}^2\right){\left(4.2\text{s}\right)}^2\\ =\left(48.988\text{m}-86.436\text{m}\right)\\ =37.448\text{m}\end{array}$

Therefore, the initial height $h$ is $\underset{\_}{37.5\text{m}}$

(d)

To determine

The maximum height $H$ that reached by the stone.

Answer to Problem 47P

The maximum height $H$ that reached by the stone is $\underset{\_}{44.44\text{m}}$ above the ground.

Explanation of Solution

Write the expression relating equation of motion.

    $v_{fy}^2=v_{iy}^2+2a_{y}y$        (V)

Conclusion:

Substitute $0$ for $v_{fy}$, $11.664\text{m}/\text{s}$ for $v_{iy}$ and $-9.8\text{m}/{\text{s}}^2$ for $a_y$ in equation (V) to find $y$.

    $\begin{array}{c}{\left(0\text{m}/\text{s}\right)}^2={\left(11.64\text{m}/\text{s}\right)}^2+2\left(-9.8\text{m}/{\text{s}}^2\right)y\\ y=\frac{{\left(11.64\text{m}/\text{s}\right)}^2}{2\left(9.8\text{m}/{\text{s}}^2\right)}\\ =6.94\text{m}\end{array}$

The maximum height reached by the stone is ,

    $\begin{array}{c}H=h+y\\ =6.94\text{m}+37.5\text{m}\\ \text{=44}\text{.44}\text{m}\end{array}$

Therefore, The maximum height $H$ that reached by the stone is $\underset{\_}{44.44\text{m}}$ above the ground.