Chapter 3, Problem 19P

A student stands at the edge of a cliff and throws a stone horizontally over the edge with a speed of v_i= 18.0 m/s. The cliff is h = 50.0 m above a body of water as shown in Figure P3.19. (a) What are the coordinates of the initial position of the stone? (b) What are the components of the initial velocity of the stone? (c) What is the appropriate analysis model for the vertical motion of the stone? (d) What is the appropriate analysis model for the horizontal motion of the stone? (e) Write symbolic equations for the x and y components of the velocity of the stone as a function of time. (f) Write symbolic equations for the position of the stone as a function of time. (g) How long after being released does the stone strike the water below the cliff? (h) With what speed and angle of impact does the stone land?

To determine

The coordinates of the initial position of the stone .

Answer to Problem 19P

The coordinates of the initial position of the stone are $\underset{\_}{x_i=0.00\text{m},y_i=0.00\text{m}}$.

Explanation of Solution

Write the expression for the initial position of the stone.

    $r_i=x_i+y_i$        (I)

Here, $r_i$ is the initial position of the stone, $x_i$ is the horizontal initial position of the stone and $y_i$ is the vertical initial position of the stone.

Conclusion:

Substitute $0.00\text{m}$ for $x_i$ and $0.00\text{m}$ for $y_i$ in (I) to find $r_i$

    $r_i=0.00\text{m}+0.00\text{m}$

Therefore, the coordinates of the initial position of the stone are $\underset{\_}{x_i=0.00\text{m},y_i=0.00\text{m}}$.

To determine

The components of the initial velocity of the stone .

Answer to Problem 19P

The components of the initial velocity of the stone are $\underset{\_}{v_{xi}=18.0\text{m/s,}{v}_{yi}=0.0\text{m/s}}$.

Explanation of Solution

Write the expression for the initial velocity of the stone,

    $v_i=v_{xi}+v_{yi}$        (II)

Here, $v_i$ is the initial velocity of the stone, $v_{xi}$ is the horizontal component of the initial velocity and $v_{yi}$ is the vertical component of the initial velocity of the stone.

Conclusion:

Substitute $18.0\text{m/s}$ for $v_{xi}$ and $0.0\text{m/s}$ for $v_{yi}$ in (II) to find $v_i$ ,

    $v_i=18.0\text{m/s}+0.0\text{m/s}$

Therefore, the components of the initial velocity of the stone are $\underset{\_}{v_{xi}=18.0\text{m/s,}{v}_{yi}=0.0\text{m/s}}$.

To determine

The vertical motion of the stone .

Answer to Problem 19P

The vertical motion of the stone is $\underset{\_}{\text{9}\text{.8}{\text{m/s}}^2}$.

Explanation of Solution

In this case, the vertical motion of the stone is equal to the free fall motion.

It is with a constant downward acceleration.

    $a_y=g$        (III)

Here, $a_y$ is the vertical downward motion and $g$ is the acceleration due to gravity.

Conclusion:

Substitute $9.8{\text{m/s}}^{\text{2}}$ for $g$ in (III) to find $a_y$,

    $a_y=9.8{\text{m/s}}^{\text{2}}$

Therefore, the vertical motion of the stone is $\underset{\_}{\text{9}\text{.8}{\text{m/s}}^2}$.

To determine

The horizontal motion of the stone .

Answer to Problem 19P

The horizontal motion of the stone is $\underset{\_}{\text{constant}\text{motion}}$.

Explanation of Solution

In this case, the constant velocity motion in the horizontal direction.

Conclusion:

Thus, there is no horizontal acceleration from gravity.

Therefore, the horizontal motion of the stone is $\underset{\_}{\text{constant}\text{motion}}$.

To determine

The symbolic equations for the $x\text{and}y$ components of the velocity of the stone as a function of time .

Answer to Problem 19P

The symbolic equations for the $x\text{and}y$ components of the velocity of the stone as a function of time are $\underset{\_}{v_{xf}=v_{xi}\text{and}{v}_{yf}=-gt}$.

Explanation of Solution

Write the expression for the horizontal final velocity of the stone.

    $v_{xf}=v_{xi}+a_{x}t$        (IV)

Here, $v_{xf}$ is the horizontal final velocity of the stone, $v_{xi}$ is the horizontal initial velocity of the stone, $a_x$ is the horizontal acceleration component of the stone and $t$ is the flight time.

Write the expression for the vertical final velocity of the stone.

    $v_{yf}=v_{yi}+a_{y}t$        (V)

Here, $v_{yf}$ is the vertical final velocity of the stone, $v_{yi}$ is the vertical initial velocity of the stone, $a_y$ is the vertical acceleration component of the stone and $t$ is the flight time.

Conclusion:

Substitute $0$ for $a_x$ in (IV) to get $v_{xf}$ ,

    $v_{xf}=v_{xi}$

Substitute $0$ for $v_{yi}$ and $-g$ for $a_y$ in V) to get $v_{yf}$ ,

    $v_{yf}=-gt$

Therefore, the symbolic equations for the $x\text{and}y$ components of the velocity of the stone as a function of time are $\underset{\_}{v_{xf}=v_{xi}\text{and}{v}_{yf}=-gt}$.

To determine

The symbolic equations for the position of the stone as a function of time .

Answer to Problem 19P

The symbolic equations for the position of the stone as a function of time are $\underset{\_}{x_f=v_{xi}t\text{and}{y}_f=-\frac{1}{2}g{t}^2}$.

Explanation of Solution

Write the expression for the horizontal final position of the stone.

    $x_f=x_i+v_{xi}t+\frac{1}{2}{a}_x{t}^2$        (VI)

Here, $x_f$ is the horizontal final position of the stone, $x_i$ is the horizontal initial position of the stone, $v_{xi}$ is the horizontal initial velocity of the stone, $a_x$ is the horizontal acceleration component of the stone and $t$ is the flight time.

Write the expression for the vertical final position of the stone.

    $y_f=y_i+v_{yi}t+\frac{1}{2}{a}_y{t}^2$        (VII)

Here, $y_f$ is the vertical final position of the stone, $y_i$ is the vertical final position of the stone, $v_{yi}$ is the vertical initial velocity of the stone, $a_y$ is the vertical acceleration component of the stone and $t$ is the flight time.

Conclusion:

Substitute $0$ for $a_x$ and $0$ for $x_i$  in (VI) to get $x_f$ ,

    $x_f=v_{xi}t$

Substitute $0$ for $y_i$, $v_{yi}$ for $0$ and $-g$ for $a_y$ in (VII) to get $y_f$ ,

    $y_f=-\frac{1}{2}g{t}^2$

Therefore, the symbolic equations for the $x\text{and}y$ components of the velocity of the stone as a function of time are $\underset{\_}{x_f=v_{xi}t\text{and}{y}_f=-\frac{1}{2}g{t}^2}$.

To determine

The time of impact of the stone .

Answer to Problem 19P

The time of impact of the stone is $\underset{\_}{t=3.19\text{s}}$.

Explanation of Solution

Write the expression for the vertical final position of the stone.

    $y_f=-\frac{1}{2}g{t}^2$        (VIII)

Here, $y_f$ is the vertical final position of the stone, $g$ is the vertical acceleration of the stone and $t$ is the flight time.

Rewrite the above equation,

    $\begin{array}{c}-h=-\frac{1}{2}g{t}^2\\ t=\sqrt{\frac{2h}{g}}\end{array}$        (IX)

Conclusion:

Substitute $9.80{\text{m/s}}^{\text{2}}$ for $g$ and $50.0\text{m}$ for $h$  in (IX) to get $t$ ,

    $\begin{array}{c}t=\sqrt{\frac{2\left(50.0\text{m}\right)}{9.80{\text{m/s}}^{\text{2}}}}\\ =3.19\text{s}\end{array}$

Therefore, the time of impact of the stone is $\underset{\_}{t=3.19\text{s}}$.

To determine

The speed and angle of impact of the stone land .

Answer to Problem 19P

The speed and angle of impact of the stone land are $\underset{\_}{v_f=36.1\text{m/s}\text{and}\theta =-{60.1}^{\text{o}}}$.

Explanation of Solution

In this case, at the time of impact $v_{xf}=v_{xi}$ .

The vertical component velocity of the stone, $v_{yf}=-gt$ .

Substitute $\sqrt{\frac{2h}{g}}$ for $t$ in the above equation,

    $\begin{array}{c}{v}_{yf}=-g\sqrt{\frac{2h}{g}}\\ =-\sqrt{2gh}\end{array}$        (X)

Write the expression for the final velocity of the stone.

    $v_f=\sqrt{v_{xf}^2+v_{yf}^2}$        (XI)

Here, $v_f$ is the final velocity of the stone

Write the expression for the angle of impact of the stone.

    $\theta ={\tan }^{-1}\frac{v_{yf}}{v_{xf}}$        (XII)

Here, $\theta$ is the angle of impact of the stone.

Conclusion:

Substitute $9.8{\text{m/s}}^{\text{2}}$ for $g$ and $50.0\text{m}$ for $h$  in (X) to get $v_{yf}$ ,

    $\begin{array}{c}{v}_{yf}=-\sqrt{2\left(9.8{\text{m/s}}^{\text{2}}\right)\left(50.0\text{m}\right)}\\ =-31.3\text{m/s}\end{array}$ .

Substitute $18.0{\text{m/s}}^{\text{2}}$ for $v_{xf}$ and $-31.3\text{m/s}$ for $v_{yf}$ in (XI) to get $v_f$ ,

    $\begin{array}{c}{v}_f=\sqrt{{\left(18.0{\text{m/s}}^{\text{2}}\right)}^2+{\left(-31.3\text{m/s}\right)}^2}\\ =36.1\text{m/s}\end{array}$ .

Substitute $-31.3\text{m/s}$ for $v_{yf}$ and $18.0\text{m/s}$ for $v_{xf}$ in (XII) to find ${\theta }_f$ ,

    $\begin{array}{c}{\theta }_f={\tan }^{-1}\frac{31.3}{18.0}\\ =-{60.1}^{\text{o}}\end{array}$

Therefore, the speed and angle of impact of the stone land are $\underset{\_}{v_f=36.1\text{m/s}\text{and}\theta =-{60.1}^{\text{o}}}$.