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

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 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.

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

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?

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hage567
Homework Helper
Saplingg, I'll send you a PM.

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.

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?