# Simple Problem with Gravity and Time

**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,

v_{1} = 0

g = 9.8 m/s^{2}

Δt = 3.0 s

v_{sound} = 332 m/s

**2. Relevant equations**

FOR SOUND

Δd = v_{sound} * Δt_{2}, where Δt_{2} is the time it takes from the sound to reach the top of the cliff from the bottom.

FOR ROCK

Δd = v_{1} * Δt + 0.5 * g * (Δt)^{2}

*the following equations may be useful but i doubt it*

v_{2} = v_{1} + g * Δt

(v_{2})^{2} = (v_{1})^{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.

http://ift.tt/1jN8YNM

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