# Math Help - Physics problem on mechanics

1. ## Physics problem on mechanics

I have a following problem

A rock is dropped from a sea cliff. After 3.4 s a splash of the rock hitting the ocean is heard. The speed of sound is 340 m/s. How tall is the cliff?

I came to the conclusion that the lenght of the cliff equals the lenght of the distance the sound travels, and that the time the rock travel and the time the sound travels should equal 3.4 seconds, but I don't know whether this is correct and how to write this in symbols.

Can someone please help me?

Thank you!

2. Originally Posted by Coach
I have a following problem

A rock is dropped from a sea cliff. After 3.4 s a splash of the rock hitting the ocean is heard. The speed of sound is 340 m/s. How tall is the cliff?

I came to the conclusion that the lenght of the cliff equals the lenght of the distance the sound travels, and that the time the rock travel and the time the sound travels should equal 3.4 seconds, but I don't know whether this is correct and how to write this in symbols.

Can someone please help me?

Thank you!
Hello,

let h be the height of the cliff
let x be the time the rock needs to fall down
then (3.4 - x) is the time the sound needs to travel up

$h = \frac12 \cdot 9.81 \cdot x^2$

$h = (3.4-x) \cdot 340$

Therefore:

$\frac12 \cdot 9.81 \cdot x^2 = (3.4-x) \cdot 340$

Solve for x.

EDIT: For confirmation only: I've got x = 3.2478 s

3. Thank you very much!

I just have one question still, do you get $\frac{1}{2}$ from a formula?

4. Originally Posted by Coach
I just have one question still, do you get $\frac{1}{2}$ from a formula?
Hej,

when the rock is falling down from the top of the cliff it's motion is a uniformly accelerated movement.

If
- a is the acceleration
- v is the speed
- d is the distance
- t is the time

you have:

v = a * t
d = 1/2 * a * t²

With your problem a = 9.81 m/s². So if you use the length of the cliff = distance of the fall you have to use the formula for d which contains the factor (1/2).