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## Homework Statement

A rock is thrown horizontally off a 56 m high cliff overlooking the ocean. And the sound of the splash is heard 3.60 seconds later. Speed of sound in air is 343 m/s.

What was the initial velocity of the rock?

2. Homework Equations

2. Homework Equations

## The Attempt at a Solution

Okay so I know

height of cliff is [itex]56m [/itex]

[itex] V_0y = 0, V_x = V_{0x} [/itex]

##t = t_1 + t_2=3.60 s##

Time it took for the rock to hit the ground I use this eq because ##V_oy = 0##

##\Delta y = V_{0y} t + \frac 12 a_x t^2##

## \sqrt {{2\Delta y }\over{a_x}} = t_1##

plugging my values in I got

##t_1 = 3.38## whichc gives ##t_2 = 0.22 ##

Since the cliff is 56 m high, I don't know the horizontal distance, but I do know that the speed of sound is traveling back in a straight line at ##343 m/s##, and with this, I multiply ##343 m/s## by ##0.22 s##, and get ##75 m##. This helps me get horizontal distance of ##50 m##. Does everyone agree with my logic here?

So now I use formula

since ##V_{0x} = V_x##

##\Delta x = {(V_{0x} + V_x)t\over 2}##

##{{\Delta x}\over t} = V_x##

plugging in I get ##14.8 m/s ##

But my book gives me answer ##15.1 m/s##

Now I believe this to be an error that has to do with rounding or significant figures, but I'm not sure, and I don't see my professor until Saturday so I can't ask him if my method is correct. Can anyone please help me here?

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