Question

A student stands at the edge of a cliff and throws a stone horizontally over the edge with a speed of${\mathit{v}}_{\mathbf{i}}\mathbf{=}\mathbf{18}\mathbf{.}\mathbf{0}\mathbf{ }\mathbf{\text{m/s}}$. The cliff is$\mathit{h}\mathbf{=}\mathbf{50}\mathbf{.}\mathbf{0}\mathbf{ }\mathbf{\text{m}}$above a body of water as shown in Figure P4.29.

(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?

Short answer

1. Coordinates for the initial position of the stone is $x_i=0 \text{m}$and$y_i=0 \text{m}$ .
2. Components of initial velocity of the stone is$u_x=v_i=18 \text{m/s}$and$u_y=0$ .
3. Stone’s vertical motion can be analysed as a free fall motion with constant downward acceleration due to gravity, $g=9.8 {\text{m/s}}^2$.
4. Stone’s horizontal motion can be analysed as a motion with a constant horizontal velocity, since there is nothing accelerating or decelerating the object in the horizontal direction.
5. The symbolic equation for the x and y components of the velocity is,$v_x=u_x$ and $v_y=-gt$.
6. The symbolic equation for the position is,$x=u_{x}t$and$y=-\frac{1}{2}g{t}^2$ .
7. Stone strikes the water below the cliff after$3.194 \text{s}$.
8. The velocity and angle are, $36.12 \text{m/s}$and $-60.098°$.

Step-by-step solution

Identification of the given data

The initial velocity is, $v_i=18.0 \text{m/s}$.

The cliff height is, $h=50.0 \text{m}$.

Significance of the horizontal velocity

When solving velocity problems in physics, the motion is split into its vertical and horizontal components. For issues involving a trajectory's angle, you employ vertical velocity. For items traveling horizontally, horizontal velocity becomes crucial.

Step 3(a): Determination of the coordinates of the initial position of the stone

Coordinates for the initial position of the stone is $x_i=0 \text{m}$and $y_i=0 \text{m}$.

Step 4(b): Determination of the components of the initial velocity of the stone

Components of initial velocity of the stone is $u_x=v_i=18\text{m/s}$and$u_y=0$ .

Step 5(c): Determination of the appropriate analysis model for the vertical motion of the stone

Stone’s vertical motion can be analysed as a free fall motion with constant downward acceleration due to gravity, $g=9.8 {\text{m/s}}^{\text{2}}$.

Step 6(d): Determination of the appropriate analysis model for the horizontal motion of the stone

Stone’s horizontal motion can be analysed as a motion with a constant horizontal velocity, since there is nothing accelerating or decelerating the object in the horizontal direction.

Step 7(e): To write the symbolic equations for the x and y components of the velocity of the stone as a function of time (e)

The relation between velocity, acceleration and time is expressed as,

$v_x=u_x+a_{x}t$

At initial $a_x=0$, so

$v_x=u_x$

And, in vertical direction, the initial velocity in vertical direction is zero so,

$$\begin{gathered} v_y=u_y+a_{y}t \\ {v}_y=-gt \end{gathered}$$

Hence the symbolic equation for the x and y components of the velocity is, $v_x=u_x$and$v_y=-gt$ .

Step 8(f): To write down the symbolic equations for the position of the stone as a function of time (f)

The relation between velocity, acceleration, distance and time is expressed as,

$x=x_i+u_{x}t+\frac{1}{2}{a}_x{t}^2$

At initial $x_i=0$and $a_x=0$, so

$x=u_{x}t$

And, at vertical direction in initial,$y_i=0$, $u_y=0$, and $a_y=-g$. So,

$$\begin{gathered} y=y_i+u_{y}t+\frac{1}{2}{a}_y{t}^2 \\ y=-\frac{1}{2}g{t}^2 \end{gathered}$$

Hence the symbolic equation for the position is, $x=u_{x}t$and $y=-\frac{1}{2}g{t}^2$.

Step 9(g): Calculating time t from the above equation (g)

Here,$y=-h$

Therefore,

$\begin{array}{rcl}-h& =& -\frac{1}{2}g{t}^2\\ t^2& =& \frac{2h}{g}\\ t& =& \sqrt{\frac{2h}{g}}\\ & & \end{array}$

Substitute all the value in the above equation.

$$\begin{gathered} t=\sqrt{\frac{2(50 \text{m})}{9.8 {\text{m/s}}^2}} \\ t=3.194 \text{s} \end{gathered}$$

Therefore, after being released, the stone strikes the water after$t=3.194 \text{s}$.

Step 10(h): Solving for the final velocity and angle of impact (h)

From (e), we got $v_x=u_x=18 \text{m/s}$i.e. Final velocity in x direction is equal to the initial velocity in x direction.

And we got$v_y=-gt$i.e. Final velocity in y direction.

Therefore,

$$\begin{gathered} v_y=\left(-9.8 {\text{m/s}}^2\right)\times \left(3.194 \text{s}\right) \\ {v}_y=-31.301 \text{m/s} \end{gathered}$$

Since,

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

Substitute all the value in the above equation.

$$\begin{gathered} v_f=\sqrt{{\left(18 \text{m/s}\right)}^2+{\left(-31.301 \text{m/s}\right)}^2} \\ {v}_f=36.12 \text{m/s} \end{gathered}$$

And the angle of impact

$$\begin{gathered} \theta ={\tan }^{-1}\left(\frac{v_y}{v_x}\right) \\ \theta ={\tan }^{-1}\left(\frac{-31.301 \text{m/s}}{18 \text{m/s}}\right) \\ \theta =-60.{098}^0 \end{gathered}$$

Hence the velocity and angle are, 36.12 m/s and$-60.{098}^0$ .