Chapter 3, Problem 23P

A student stands at the edge of a cliff and throws a stone horizontally over the edge with a speed of 18.0 m/s. The cliff is 50.0 m above a flat, horizontal beach as shown in Figure P3.7. (a) What are the coordinates of the initial position of the stone? (b) What are the components of the initial velocity? (c) Write the equations for the x- and y-components of the velocity of the stone with time, (d) Write the equations for the position of the stone with time, using the coordinates in Figure P3.7. (e) How long after being released does the stone strike the beach below the cliff? (f) With what speed and angle of impact does the stone land?

Figure P3.7

To determine

The coordinates of the initial position of the stone.

Answer to Problem 23P

The coordinates of the initial position of the stone are $x_0=0\text{m}$ and $y_0=50.0\text{m}$.

Explanation of Solution

Initially, as the student is standing at the top of the cliff, the $$ coordinate of the stone is zero and the $y$ coordinate of the stone will be equal to the height of the cliff.

Conclusion:

Thus, the coordinates of the initial position of the stone are $x_0=0\text{m}$ and $y_0=50.0\text{m}$.

To determine

The components of the initial velocity of the stone.

Answer to Problem 23P

The coordinates of the initial velocity of the stone are $v_{0x}=18.0\text{m/s}$ and $v_{0y}=0\text{m/s}$.

Explanation of Solution

When the stone is thrown from the top of the cliff, the case is of a horizontally launched projectile. For a horizontally launched projectile, initially, the velocity is perpendicular to the acceleration.

The velocity is directed in the horizontal direction and the velocity in the vertical direction is zero.

Conclusion:

For the horizontally launched projectile, the initial velocity with which the stone is thrown is directed in the horizontal direction and the velocity in the vertical direction is zero.

The coordinates of the initial velocity of the stone are $v_{0x}=18.0\text{m/s}$ and $v_{0y}=0\text{m/s}$.

To determine

The equations for $x$ and $y$ components of the velocity of the stone with time.

Answer to Problem 23P

The equations for $x$ and $y$ components of the velocity of the stone with time are $v_x=18.0\text{m/s}$ and $v_y=-\left(9.80{\text{m/s}}^2\right)t$.

Explanation of Solution

The components of the velocity of the stone can be found from the kinematic equations of motion.

The kinematic equations of motion for the stone are

$\begin{array}{l}v=v_0+at\\ s=s_0+v_{0}t+\frac{1}{2}a{t}^2\\ v^2-v_0^2=2as\end{array}$

Here,

$v_0$ is the initial velocity

$v$ is the final velocity

$s_0$ is the initial position

$s$ is the displacement

$a$ is the acceleration

$t$ is the time taken

Conclusion:

The equations for the components of the velocity can be obtained a function of time.

The velocity is in the horizontal direction and the acceleration in the horizontal direction is zero.

In the horizontal direction, the first equation becomes

$v_x=v_{0x}+a_{x}t$

Substitute $18.0\text{m/s}$ for $v_{0x}$ and $0{\text{m/s}}^2$ for $a_x$.

$\begin{array}{c}{v}_x=18.0\text{m/s}+\left(0{\text{m/s}}^2\right)t\\ =18.0\text{m/s}\end{array}$

The velocity is in the vertical direction and the acceleration in the vertical direction is the acceleration due to gravity.

In the vertical direction, the equation for velocity becomes

$v_y=v_{0y}+a_{y}t$

Substitute $0\text{m/s}$ for $v_{0y}$ and $-9.8{\text{m/s}}^2$ for $a_y$.

$\begin{array}{c}{v}_y=0\text{m/s}+\left(-9.8{\text{m/s}}^2\right)t\\ =\left(-9.8{\text{m/s}}^2\right)t\end{array}$

Thus, the equations for $x$ and $y$ components of the velocity of the stone with time are $v_x=18.0\text{m/s}$ and $v_y=-\left(9.80{\text{m/s}}^2\right)t$.

To determine

The equations for the position of the stone with time.

Answer to Problem 23P

The equations for the position of the stone with time are $x=\left(18.0\text{m/s}\right)t$ and $y=50.0\text{m}-\left(4.90{\text{m/s}}^2\right)t^2$.

Explanation of Solution

The kinematic equations of motion for the stone are

$\begin{array}{l}v=v_0+at\\ s=s_0+v_{0}t+\frac{1}{2}a{t}^2\\ v^2-v_0^2=2as\end{array}$

Here,

$v_0$ is the initial velocity

$v$ is the final velocity

$s_0$ is the initial position

$s$ is the displacement

$a$ is the acceleration

$t$ is the time taken

Conclusion:

In the horizontal direction, the equation for position becomes

$x=x_0+v_{0x}t+\frac{1}{2}{a}_x{t}^2$

Substitute $0\text{m}$ for $x_0$, $18.0\text{m/s}$ for $v_{0x}$ and $0{\text{m/s}}^2$ for $a_x$.

$\begin{array}{c}x=0\text{m}+\left(18.0\text{m/s}\right)t+\frac{1}{2}\left(0{\text{m/s}}^2\right)t^2\\ =\left(18.0\text{m/s}\right)t\end{array}$

In the vertical direction, the equation for position becomes

$x=x_0+v_{0x}t+\frac{1}{2}{a}_x{t}^2$

Substitute $50.0\text{m}$ for $y_0$, $0\text{m/s}$ for $v_{0y}$ and $-9.8{\text{m/s}}^2$ for $a_y$.

$\begin{array}{c}y=50.0\text{m}+\left(0\text{m/s}\right)t+\frac{1}{2}\left(-9.8{\text{m/s}}^2\right)t^2\\ =50.0\text{m}-\left(4.9{\text{m/s}}^2\right)t^2\end{array}$

Thus, the equations for the position of the stone with time are $x=\left(18.0\text{m/s}\right)t$ and $y=50.0\text{m}-\left(4.90{\text{m/s}}^2\right)t^2$.

To determine

The time taken for the stone to strike the beach.

Answer to Problem 23P

The time taken for the stone to strike the beach is $3.19\text{s}$.

Explanation of Solution

The displacement in the vertical direction can be written as

$\Delta y=v_{0y}t+\frac{1}{2}{a}_y{t}^2$

Initially, as the velocity in the vertical direction is zero, the equation becomes

$\Delta y=\frac{1}{2}{a}_y{t}^2$

So, the equation for time taken is,

$t=\sqrt{\frac{2\left(\Delta y\right)}{a}}$

Conclusion:

The equation for the time taken for the stone to reach the beach is,

$t=\sqrt{\frac{2\left(\Delta y\right)}{a}}$

Substitute $\left(-50.0\text{m}\right)$ for $\Delta y$ and $-9.80{\text{m/s}}^2$ for $a$.

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

Thus, the time taken for the stone to strike the beach is $3.19\text{s}$.

To determine

The speed and angle of impact of the stone when it lands.

Answer to Problem 23P

The speed and angle of impact of the stone when it lands are $\overrightarrow{v}=36.1\text{m/s}\text{at}60.1°\text{below the horizontal}$.

Explanation of Solution

The horizontal velocity of the projectile motion is constant throughout the motion.

Thus, $v_x=v_{0x}=18.0\text{m/s}$

The velocity in the vertical direction is found from the equation

$v_y=v_{0y}+a_{y}t$

Substitute $0\text{m/s}$ for $v_{0y}$, $\left(-9.8{\text{m/s}}^2\right)$ for $a_y$ and $3.19\text{s}$ for $t$.

$\begin{array}{c}{v}_y=0\text{m/s}+\left(-9.80{\text{m/s}}^2\right)\left(3.19\text{s}\right)\\ =-31.1\text{m/s}\end{array}$

Conclusion:

The magnitude of the velocity can be found from the horizontal and vertical components of the velocity.

The velocity is given by

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

Substitute $18.0\text{m/s}$ for $v_x$ and $-31.3\text{m/s}$ for $v_y$.

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

The direction of the velocity is found from the relation

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

Substitute $18.0\text{m/s}$ for $v_x$ and $-31.3\text{m/s}$ for $v_y$.

$\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{-31.3\text{m/s}}{18.0\text{m/s}}\right)\\ =-60.1°\end{array}$

Thus, the speed and angle of impact of the stone when it lands are $\overrightarrow{v}=36.1\text{m/s}\text{at}60.1°\text{below the horizontal}$.