A rock is dropped from a sea cliff, and the sound of it striking the ocean is heard 3.2 s later. If the speed of sound is 340 m/s, how high is the cliff?
vo= 0; s=340 m/s; t= 3.2s
h=?
t1+t2= 3.2 s
t2= 3.2-t1
I used Eq: h= vo*t^1+1/2at^2
rearranged eq.
h= (4.9)(t)^2
t^2= h/4.9
t^2= 340/4.9= 69.38 s
substitute--> 69.38=3.2-t^1
t^1= -66.18 s
Sorry, I don't know if I am doing this right.I still can't figure out how to get h=?
d=1/2at^2
t=3.2s (this time includes both the time taken for the rock to fall and the time taken by sound to travel to the listener from the ocean)
from s=d/t
t=d/s
change in time=t-d/s
therefore d=1/2a(t-d/s)^2
d=1/2(9.8)(3.2-d/340)^2
work this out and solve for d, which is the height of the cliff
To solve for the height of the cliff, we can use the equation of motion for an object in free fall.
The equation is:
h = vo * t + (1/2) * a * t^2
In this case, the initial velocity (vo) is 0 since the rock is dropped. The acceleration (a) is the acceleration due to gravity, which is approximately 9.8 m/s^2.
So the equation becomes:
h = (1/2) * a * t^2
We can substitute the given values into the equation:
h = (1/2) * (9.8 m/s^2) * (3.2 s)^2
Now, let's calculate it step by step:
First, square the time: (3.2 s)^2 = 10.24 s^2
Next, multiply by the acceleration due to gravity: (9.8 m/s^2) * (10.24 s^2) = 100.352 m
Therefore, the height of the cliff is approximately 100.352 meters.