A rock is dropped from a sea cliff, and the sound of it striking the ocean is heard 3.3 later. If the speed of sound is 340 , how high is the cliff?
Given:
t=5sec.
Vo=0m/s
g=9.81m/s^2
V=320m/s^2
Solution:
H=Vot+0.5gt^2
H=(0.5)(9.81)(5^2)
H=122.625m
H=vt
H=320(5)
H=1600m
H=122.625+1600
[H=1722.625m] :)
You need to show the units of your numbers. I will assume they are meters and seconds.
Solve this equation for height, H:
3.23 s = time for sound to arrive + time to fall
3.3 = H/340 + sqrt(2H/9.8)
Given:
t=3.3sec.
Vo=0m/s
g=9.81m/s^2
V=340m/s^2
Solution:
H=Vot+0.5gt^2
H=(0.5)(9.81)(3.3^2)
H=53.42m
H=vt
H=340(5)
H=1700m
H=53.42+1700
[H=1753.42m] :)
1.3 (using the given equation...)
To determine the height of the cliff, we need to first calculate the time it takes for the rock to fall and then use that time to find the height.
We know that the speed of sound is 340 m/s and that the sound of the rock reaching the ocean is heard 3.3 seconds later. This means that the sound took 3.3 seconds to travel from the rock to our ears.
Now, let's calculate the time it took for the rock to fall. We can assume there is no initial vertical velocity, so the only force acting on the rock is gravity. We can use the equation for free fall:
s = ut + (1/2)gt^2
Where:
s = distance (height of the cliff)
u = initial velocity (0 m/s)
g = acceleration due to gravity (approximately -9.8 m/s², taking into account the downward direction)
t = time
Since the rock was dropped, the initial velocity is zero, and the equation simplifies to:
s = (1/2)gt^2
Plugging in the values we know:
s = (1/2)(-9.8)(3.3)^2
Calculating this expression will give us the height of the cliff.