Chapter 2, Problem 45P

To determine

The height of cliff.

Answer to Problem 45P

46 m.

Explanation of Solution

Given:

Initial velocity of the rock as it dropped from the cliff $=v_i=0\frac{\text{m}}{\text{s}}$

Time taken by rock to strike the ocean $=t_r$

Acceleration experienced by rock = acceleration due to gravity $=a=-9.8\frac{\text{m}}{{\text{s}}^{\text{2}}}$

Height of the cliff $=h$

Initial position of rock as it leaves the cliff $=y_i=h$

Final position of rock as it strikes the ocean $=y_f=0\text{m}$

Speed of sound $=v_s=340\frac{\text{m}}{\text{s}}$

Distance traveled by sound after rock hits the ocean $=d=h$

Time taken by sound to travel from ocean to top of cliff $=t_s$

Time of hearing the sound of rock striking the ocean $=t=3.2\sec$

Formula Used:

Kinematics equation that relates initial position, final position, initial velocity, acceleration and time of an object is given as

  $y_f=y_i+v_{i}t+(0.5)a{t}^2$

For a constant speed motion, distance is related to time and speed as

  $d=s\times t$

Calculation:

Consider the motion of sound after rock strikes the ocean.

Using the equation

  $\begin{array}{l}d=v_s{t}_s\\ h=v_s{t}_s\text{ since }d=h\\ h=(340)t_s\text{ since }{v}_s=340\\ t_s=\frac{h}{340}\left(1\right)\end{array}$

Total time for hearing the sound of rock striking the ocean is given as

  $\begin{array}{l}t=t_r+t_s\\ 3.2=t_r+t_s\end{array}$

Using equation (1).

  $\begin{array}{l}3.2=t_r+\left(\frac{h}{340}\right)\\ t_r=3.2-\left(\frac{h}{340}\right)\left(2\right)\end{array}$

Consider the motion of the rock as it falls from the cliff.

Use the kinematics equation

  $y_f=y_i+v_{i}t+(0.5)a{t}^2$

Substituting the values

  $\begin{array}{l}0=h+(0)t_r+(0.5)(-9.8)t_r^2\\ 0=h-(4.9)t_r^2\end{array}$

Using equation (2)

  $\begin{array}{l}0=h-(4.9){\left(3.2-\left(\frac{h}{340}\right)\right)}^2\\ h=46\text{m}\end{array}$

Conclusion:

Therefore, height of the cliff is 46 m.