Answer:
height of cliff (h) = 112.38m
Step-by-step explanation:
The time 5.12 s is the total time it takes for the rock to fall, and for the soundwave to travel back to the top of the cliff before it is heard.
![5.12 = t_f\ +\ t_s - - - - -(1)\\where:\\t_f = time\ of\ fall\ of\ the\ piece\ of\ rock\\t_s = time\ travelled\ by\ the\ return\ sound](https://img.qammunity.org/2021/formulas/physics/college/8v8wg4fqj265uehw8oop44yu9wy3uern7u.png)
Let h be the height of the cliff in meters, the time taken for the rock to fall is given by:
![t_f=\sqrt{(2h)/(g) } \\where:\\t_f = time\ of\ fall\\h = height\ of\ cliff\\g= acceleration\ due\ to\ gravity= 9.8 m \slash s^2](https://img.qammunity.org/2021/formulas/physics/college/xznt1u23a14ywh8nlwxrrx3k20qem5h45v.png)
![\therefore t_f = \sqrt{(2h)/(9.8)} \\squaring\ both\ sides\\(t_f)^2 = (2h)/(9.8)\\ 2h = 9.8 *s\ (t_f)^2\\h = (9.8 *s\ (t_f)^2)/(2) \\h= 4.9(t_f)^2 - - - - - (2)](https://img.qammunity.org/2021/formulas/physics/college/nlyf57vltpo5jyfti3pa1cqifcgffmprcv.png)
Next, let us calculate the time taken fot the sound to return
![t_s = (h)/(v) \\where:\\t_s = time\ for\ sound\ to\ travel\ up\ the\ cliff\\h= distance\ tavelled\ = height\ of\ cliff\\v= speed\ = 340m \slash s\ (speed\ of\ sound)\\\therefore t_s = (h)/(340) - - - - - (3)\\](https://img.qammunity.org/2021/formulas/physics/college/4hn5ow19qmy228hpze86f8is9ktp1ip3je.png)
now putting the values of h from equation 2 into equation (3)
![t_s = (4.9(t_f)^2)/(340)](https://img.qammunity.org/2021/formulas/physics/college/gh85hohorklfoby4qnbtdwefvxnjylaybi.png)
Putting the value of
into equation (1)
![5.12 = t_f +(4.9(t_f)^2)/(340) \\](https://img.qammunity.org/2021/formulas/physics/college/ncvhtw5acefio8tga8v799zlzji27od2fc.png)
multiplying through by 340
![1740.8 = 340(t_f) + 4.9(t_f)^2\\4.9(t_f)^2 + 340 (t_f) - 1740.8 = 0](https://img.qammunity.org/2021/formulas/physics/college/c4te5wjawuzfz8k7400rlq8wj5i93ffw6u.png)
now let us solve the quadratic equationsss;
![Let\ (t_f) = x](https://img.qammunity.org/2021/formulas/physics/college/rm8dg2z1epk03hhgf90g7kzvlcrp1omu7m.png)
![4.9x^2 + 340x - 1740.8 = 0\\using\ quadratic\ formula\\x = (-b \pm√(b^2 - 4ac) )/(2a) \\x = (-340 \pm√((340)^2 -\ 4 *4.9 *(-1740.8)) )/(2*4.9)\\x = (-340\ \pm\ 386.936)/(9.8) \\x =(386.938 - 340)/(9.8) \\x = (46.936)/(9.8)\\ x = 4.789\\x = t_f\\t_f=4.789s](https://img.qammunity.org/2021/formulas/physics/college/7byhcsohwtltg49k0f7zd1y6xw4wq72max.png)
note, time cannot be negative, so we ignored the negative answer
putting the value of
into equation (2) to find height of cliff (h)
![h= 4.9(t_f)^2\\h = 4.9 *(4.789)^2\\h = 112.38m](https://img.qammunity.org/2021/formulas/physics/college/3c1ggxvwld3lz13d6gj63s62f89z3g1i8m.png)
Therefore, height of cliff (h) = 112.38m