If the speed of sound is 330 m/s. How high is the cliff?

[galeontiger]

Hello, I tried solving this but to no avail.
Maybe someone could help me figure out the answer, thanks.

A rock is dropped from a sea cliff and the sound is heard striking the ocean 3.0s later. If the speed of sound is 330 m/s. How high is the cliff?

Thanks.


I've already tried about everything that I knew.
I know that the time it takes the rock to fall to the bottom and the time it takes for the sound to travel back up to the person takes a total of 3s.

I tried using the d=v1t + 1/2at^2 formula but that didn't work either.

I've thought of and tried to use most formulas, but I don't know how to do this problem. Please help me.

 

[d_leet]

What have you tried so far? You need to show some work, and if this is homework it really belongs in the homework forum.

 

[saplingg]

Just remember that the time taken for the sound to reach your ears = 3 - the time taken for the rock to hit the water, and solve simultaneously.

@hage: How is my answer wrong?

 

[hage567]

Saplingg, It is against PF rules to provide complete solutions to homework problems.

Please read the guidelines: https://www.physicsforums.com/showthread.php?t=5374

I don't think your answer is right anyway.

 

[hage567]

Saplingg, I'll send you a PM.

 

[drpizza]

Hmmmm... I wonder if the problem is expressed incorrectly or if a teacher is just trying to make a point. Since you only have 2 significant digits for the time, the amount of time it takes makes no significant difference. In fact, using 3.00s for the time, and 9.81m/s^2, as well as 330m/s for the speed of sound, it still results in the same answer after rounding to three significant digits.

 

[heidari]

I think if you drop the rock, you can assume that 3 second is the time the rock reaches the sea plus the time is needed for the sound to reach your ear so if we consider h as the height
3= h/330+t
and for t we have: h=1/2gt^2
if we replace the t in the first equation from the second one
we can solve it!
is it ok?