Chapter 4, Problem 4.29P

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 P4.13. (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?

Figure P4.13

To determine

The coordinates of the initial position of the stone.

Answer to Problem 4.29P

The coordinates of the initial position of the stone is $\left(0,50.0\text{m}\right)$.

Explanation of Solution

The speed of stone is $18.0\text{m}/\text{s}$ and the height of cliff is $50.0\text{m}$.

Write the expression for the initial x coordinate of stone

    $x_{\text{i}}=0$

Here, $x_{\text{i}}$ is the initial x-coordinate of the stone.

Write the expression for the initial y coordinate at the top of the cliff

    $y_{\text{i}}=50\text{ m}$

Here, $y_{\text{i}}$ is the initial y-coordinate of the stone.

Conclusion:

Therefore, the coordinates of the initial position of the stone is $\left(0,50.0\text{m}\right)$.

To determine

The components of the initial velocity of the stone.

Answer to Problem 4.29P

The components of the initial velocity of the stone are $v_{\text{xi}}=18.0\text{m}/\text{s}$ and $v_{\text{yi}}=0\text{m}/\text{s}$.

Explanation of Solution

Write the expression for the initial velocity of stone in x direction

    $v_{\text{ix}}=18.0\text{m}/\text{s}$

Here, $v_{\text{i x}}$ is the initial horizontal velocity of stone.

Write the expression for the initial velocity of stone in y direction

    $v_{\text{iy}}=0\text{m}/\text{s}$

Here, $v_{\text{i y}}$ is the initial vertical velocity of stone.

Conclusion:

Therefore, the initial components of the velocity of the stone are $v_{\text{xi}}=18.0\text{m}/\text{s}$ and $v_{\text{yi}}=0\text{m}/\text{s}$.

To determine

A model that explains the vertical motion of the stone.

Answer to Problem 4.29P

In the $y$ direction the acceleration is constant and equal to the acceleration due to gravity.

Explanation of Solution

The vertical motion of the stone is governed by a free fall motion under constant acceleration $g$.

Thus, the motion in the $y$ direction is particle under constant acceleration.

Conclusion:

Therefore, in the y direction the acceleration is constant.

To determine

A model that explains the horizontal motion of the stone.

Answer to Problem 4.29P

In the x direction the velocity is constant.

Explanation of Solution

In the $x$ direction there is no net force acting to change the inertia of the stone so, there is zero acceleration in the $x$ direction so, the velocity of the particle remains constant throughout the motion.

Thus, the motion in the $y$ direction is constant velocity motion.

Conclusion:

Therefore, in the x direction, motion is due to constant velocity motion.

To determine

The symbolic equation for the x and y component of the velocity as a function of time.

Answer to Problem 4.29P

The x and y component of the velocity are $v_{\text{xf}}=v_{\text{xi}}$ and $v_{y\text{f}}=-gt$ respectively.

Explanation of Solution

In the $x$ direction no acceleration acts on the stone therefore, its speed remains constant throughout the motion.

The final velocity of stone in x direction is same as the initial velocity in the $x$ direction.

Write the relation between the final and initial velocity in the x-direction,

    $v_{\text{fx}}=v_{\text{ix}}$

Here, $v_{\text{fx}}$ is the final horizontal velocity.

The velocity in x direction does not depend on time.

In the $y$ direction a constant acceleration $g$ acts on the stone therefore, the velocity of stone in y direction.

Write the expression for final velocity in the y-direction

    $v_{\text{fy}}=v_{\text{iy}}+at$

Here, $v_{\text{fx}}$ is the final vertical velocity and $a$ is the acceleration in the vertical direction.

Conclusion:

Substitute $0$ for $v_{\text{iy}}$ and $-g$ for $a$ in the above equation to find $v_{\text{fy}}$

    $v_{\text{fy}}=-gt$

Therefore, the x and y component of the velocity are $v_{\text{xf}}=v_{\text{xi}}$ and $v_{\text{fy}}=-gt$ respectively.

To determine

The symbolic equation for the x and y component for the position as a function of time.

Answer to Problem 4.29P

The x and y component of the position are $x_{\text{f}}=v_{\text{xi}}t$ and $y_{\text{f}}=y_{\text{i}}-\frac{1}{2}g{t}^2$ respectively.

Explanation of Solution

Write the expression for the position of stone in x direction

  $x_{\text{f}}=x_{\text{i}}+v_{\text{ix}}t+a_{\text{x}}{t}^2$                                                                              (I)

Here, $x_{\text{f}}$ is the final horizontal position and $a_{\text{x}}$ is the acceleration in the x-direction.

Write the expression for the position of stone in y direction

      $y_{\text{f}}=y_{\text{i}}+v_{\text{iy}}t+\frac{1}{2}a{t}^2$                                                                 (II)

Here, $y_{\text{f}}$ is the final horizontal position.

Conclusion:

Substitute $0$ for $x_{\text{i}}$ and $0$ for $a_{\text{x}}$ in the equation (I)

  $\begin{array}{c}{x}_{\text{f}}=0+v_{\text{ix}}t+0\\ =v_{\text{ix}}t\end{array}$

Substitute $0$ for $v_{\text{iy}}$ and $-g$ for $a$ in the equation (II)

  $\begin{array}{c}{y}_{\text{f}}=y_{\text{i}}\text{+0}-\frac{1}{2}\left(g\right)t^2\\ =y_{\text{i}}-\frac{1}{2}\left(g\right)t^2\end{array}$

Therefore, the x and y component of the position are $v_{\text{xi}}t$ and $y_{\text{f}}=y_{\text{i}}-\frac{1}{2}\left(g\right)t^2$ respectively.

To determine

The time for the stone to strike the water below the cliff.

Answer to Problem 4.29P

The time for the stone to strike the water below the cliff is $3.19\text{ s}$.

Explanation of Solution

Write the formula to calculate time

    $t=\sqrt{\frac{2h}{g}}$

Conclusion:

Substitute $50\text{ m}$ for $h$ and $9.8\text{m}/{\text{s}}^2$ for $g$ in above equation to find $t$.

    $\begin{array}{c}t=\sqrt{\frac{2\left(50\text{ m}\right)}{9.8\text{m}/{\text{s}}^2}}\\ =3.19\text{ s}\end{array}$

Therefore, the time for the stone to strike the water below the cliff is $3.19\text{ s}$.

To determine

The speed and angle with which the stone land.

Answer to Problem 4.29P

The speed of the stone is $36.1\text{m}/\text{s}$ and the angle is $-60.1°$ below the horizontal.

Explanation of Solution

Write the expression to calculate velocity,

    $v_{\text{fy}}=-gt$

Substitute $3.19\text{ s}$ for $t$ and $9.8\text{m}/{\text{s}}^2$ for $g$ in above equation to find $v_{\text{fy}}$.

    $\begin{array}{c}{v}_{\text{fy}}=\left(9.8\text{m}/{\text{s}}^2\right)\left(3.19\text{ s}\right)\\ =-31.26\text{m}/\text{s}\\ \approx -31.3\text{m}/\text{s}\end{array}$

Write the formula to calculate speed of stone land

    $v=\sqrt{v_{\text{fx}}^2+v_{\text{fx}}^2}$                                                                (III)

Here, $v$ is the speed of the stone while landing.

Write the expression to calculate angle of stone land

    $\theta ={\tan }^{-1}\left(\frac{v_{\text{fy}}}{v_{\text{fx}}}\right)$                                                                (IV)

Conclusion:

Substitute $18\text{m}/\text{s}$ for $v_{\text{fx}}$ and $-31.3\text{m}/\text{s}$ for $v_{\text{fy}}$ in equation (III) to find $v$.

    $\begin{array}{c}v=\sqrt{{\left(18\text{m}/\text{s}\right)}^2+{\left(-31.3\text{m}/\text{s}\right)}^2}\\ =36.1\text{m}/\text{s}\end{array}$

Therefore, the speed of the stone at the time of land is $36.1\text{m}/\text{s}$.

Substitute $18\text{m}/\text{s}$ for $v_{\text{fx}}$ and $-31.3\text{m}/\text{s}$ for $v_{\text{fy}}$ in equation (IV) to find $\theta$.

    $\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{-31.3\text{m}/\text{s}}{18\text{m}/\text{s}}\right)\\ =-60.09°\\ \approx -60.1°\end{array}$

Therefore, the angle of the stone land is $-60.1°$ below the horizontal.