Answer:
height of cliff (h) = 112.38m
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.
[tex]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[/tex]
Let h be the height of the cliff in meters, the time taken for the rock to fall is given by:
[tex]t_f=\sqrt{\frac{2h}{g} } \\where:\\t_f = time\ of\ fall\\h = height\ of\ cliff\\g= acceleration\ due\ to\ gravity= 9.8 m \slash s^2[/tex]
[tex]\therefore t_f = \sqrt{\frac{2h}{9.8}} \\squaring\ both\ sides\\(t_f)^2 = \frac{2h}{9.8}\\ 2h = 9.8 \timess\ (t_f)^2\\h = \frac{9.8 \timess\ (t_f)^2}{2} \\h= 4.9(t_f)^2 - - - - - (2)[/tex]
Next, let us calculate the time taken fot the sound to return
[tex]t_s = \frac{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 = \frac{h}{340} - - - - - (3)\\[/tex]
now putting the values of h from equation 2 into equation (3)
[tex]t_s = \frac{4.9(t_f)^2}{340}[/tex]
Putting the value of [tex]t_s[/tex] into equation (1)
[tex]5.12 = t_f +\frac{4.9(t_f)^2}{340} \\[/tex]
multiplying through by 340
[tex]1740.8 = 340(t_f) + 4.9(t_f)^2\\4.9(t_f)^2 + 340 (t_f) - 1740.8 = 0[/tex]
now let us solve the quadratic equationsss;
[tex]Let\ (t_f) = x[/tex]
[tex]4.9x^2 + 340x - 1740.8 = 0\\using\ quadratic\ formula\\x = \frac{-b \pm\sqrt{b^2 - 4ac} }{2a} \\x = \frac{-340 \pm\sqrt{(340)^2 -\ 4 \times4.9 \times(-1740.8)} }{2\times4.9}\\x = \frac{-340\ \pm\ 386.936}{9.8} \\x =\frac{386.938 - 340}{9.8} \\x = \frac{46.936}{9.8}\\ x = 4.789\\x = t_f\\t_f=4.789s[/tex]
note, time cannot be negative, so we ignored the negative answer
putting the value of [tex]t_f[/tex] into equation (2) to find height of cliff (h)
[tex]h= 4.9(t_f)^2\\h = 4.9 \times(4.789)^2\\h = 112.38m[/tex]
Therefore, height of cliff (h) = 112.38m
