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
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=?
1 answer
1 answer