Answer:
[tex]\Delta y=221.8*m[/tex]
Step-by-step explanation:
The distance that the rock has traveled to reach the ground (also, the height of the cliff) can be calculated using:
[tex]\Delta y=y_{0}+v_{0}*t+\frac{1}{2} *g*(t_{r})^{2}[/tex]
[tex]\Delta y[/tex] is our incognit
[tex]y_{0}[/tex] is the starting position of the rock, we'll define it as 0
[tex]v_{0}[/tex] is the starting velocity of the rock, it is 0 since it starts from rest
g is the acceleration of gravity
[tex]t_{r}[/tex] is the time it took the rock reaching the ground
Also, we can model the "movement" of sound as it follows:
[tex]\Delta y=v_{s}*t_{s}[/tex]
where:
[tex]v_{s}[/tex] is the velocity of sound :330m/s
[tex]t_{s}[/tex] is the time it took the sound travelling to the top of the cliff
using the first and the second equation, and the fact that [tex]t_{total}=7.4*s=t_{r}+t_{s}[/tex] we get to:
[tex]v_{s}*(t_{total}-t_{r})=\frac{1}{2} *g*(t_{r})^{2}[/tex]
[tex]330*\frac{m}{s} *(7.4*s-t_{r})=\frac{1}{2} *9.8*\frac{m}{s^{2} } *(t_{r})^{2}[/tex]
[tex]t_{r}=6.728 *s[/tex] (we get to values for tr since the equation is quadratic, the correct one is the positive)
Now that we have [tex]t_{r}[/tex] we can use it to calculate [tex]\Delta y[/tex] and determine the height of the cliff:
[tex]\Delta y=\frac{1}{2} *9.8*\frac{m}{s^{2} } *(t_{r})^{2}[/tex]
[tex]\Delta y=\frac{1}{2} *9.8*\frac{m}{s^{2} } *(6.728*s)^{2}=221.8*m[/tex]
