What I did:

Again I drew a diagram. I let my y=-40m from where my origin is. First I calculated the sound coming back to the soccer player, as that information was given in more detail. So I know the formula X=(Vi)(t) .

So (343)(3.00) = 1024m. So I know my hypoteneuse and my opposite side which is -40. So if I apply the pith. theorm, I get:

1024^2 - (-40)^2 = 1023.2184 m

So now I know my displacement.

At this point I want the total time from the kick to when it falls into the water so that I can calculate the initial velocity. I assume that I can use this formula:

xf = vyi t -1/2 gt^2

I assume that vyi is 0 because the soccer player kicks it straight and than it goes down (don't know if this is a valid assumption). Also, I subtract t-3 since we want to see when the rock hits the water, not includng the sound back to the the soccer player.

-40 = 0 - 1/2 (9.8)(t-3)^2

-40 = -4.9 (t^2-6t+9)

-40 = -4.9t^2 = 29.4t -44.1

= 4.9t^2 - 29.4t = 4.1

since I have a quadratic equation, I use a quadratic formula which gives me a time of 5.8 seconds. which is essentially time of flight. So when when I find the intial velocity, i plug it into this equation:

x = (v)(t)

1024 = (v) (5.8)

= 176 m/s

( I don't know, but I really doubt this answer for some reason, so if someone can correct it for me, it would be much appreciated!)