# Simple Problem with Gravity and Time

1. Mar 30, 2014

### Jerry

1. The problem statement, all variables and given/known data

"A rock is dropped off a cliff into the water below. The sound of the splash is heard 3.0 s later. If the speed of sound is 332 m/s, calculate the height of the cliff above the water. (Note: the total time it takes for the rock to fall and the sound to travel upwards is 3.0 s)"

Therefore,
v1 = 0
g = 9.8 m/s2
Δt = 3.0 s
vsound = 332 m/s

2. Relevant equations
FOR SOUND
Δd = vsound * Δt2, where Δt2 is the time it takes from the sound to reach the top of the cliff from the bottom.
FOR ROCK
Δd = v1 * Δt + 0.5 * g * (Δt)2
*the following equations may be useful but i doubt it*
v2 = v1 + g * Δt
(v2)2 = (v1)2 + 2 * g * Δd

*assume no air resistance

3. The attempt at a solution
Δd = ?
No clue.

I figured that the answer should be 41 m, I believe, by trial and error but I would like to know how this can be solved in a normal way.

2. Mar 30, 2014

### Simon Bridge

The Δt for the rock is different from the Δt you set for the total time.
Call it Δt1 ?

Then Δt1+Δt2=what?

Now you have three equations and three unknowns.

3. Mar 30, 2014

### Jerry

It should be time for rock + time for sound = 3.0s and that the sound is not affected by gravity but rock is accelerating from 0 , at 9.8m/s2.

4. Mar 30, 2014

### Simon Bridge

Well done - now you can solve it.
hint: all three equations have to be true simultaneously.