How High is the Cliff if a Rock Splashes Water in 3 Seconds?

[Jerry]

Homework Statement

"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
v_sound = 332 m/s

Homework Equations

FOR SOUND
Δd = v_sound * 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

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.

 

[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.

 

[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.

 

[Simon Bridge]

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