Initial Velocity of a kicked rock

[Claymore]

Homework Statement

A soccer player kicks a rock horizontally off a 40.0-m-high cliff into a pool of water. If the player hears the sound of the splash 3.00s later, what was the initial speed given to the rock? Assume the speed of sound in air to be 343 m/s.

Homework Equations

Well to start off I have a graph which I don't know how to put it in here. To get things started I am currently taking General Physics and it seems a bit confusing than I thought.

Teacher gave us this:

r= x= ro + Vox t
y= Yo + Voy t + (gt^2)/2


v= Vfx= Vox
Vfy= Voy + gt

Im not sure if there's more to this ^

To my understanding where r= displacement if I am correct, and v is velocity(speed)

it seems that i plug in numbers and calculate them but I am having problem understanding the letters what they represent so i can't get to my solution or explain that, that's my solution. It may sound confusing

The Attempt at a Solution

Height= 40m
a=g=9.8 m/s^2
speed of sound= 342m/s
Time=3s

r = r0 + Vot + (gt^2)/2
= 0 + 342 m/s *3.0s -4.9m/s^2 * 3.0s^2
= 984.9 m

so then I used:
x = x0 Voxt
984.9m = 0 + Vox * 3s
Vox= 328.3 m/s

and:
y= y0 + Voyt + (gt^2)/2
40m= 0 + Voy*3 + -4.9m/s^2 * 3.0s^2
Voy= -28m/s

and I am stuck :( any help on what/how to get intial Velocity. Thanks in advance

Im new to forums as well :P

 

[learningphysics]

I'm finding it difficult to follow your solution... First step... what is the distance between the place where the soccer ball fell and where the soccer player is... hint: use the speed of sound.

 

[Claymore]

well that's the thing how would i find the distance? oh and by the way its a rock (ball) :P

 

[learningphysics]

well that's the thing how would i find the distance? oh and by the way its a rock (ball) :P

lol, why is he kicking a rock? hint: they give the speed of sound... they give the time it takes to hear the sound... what distance does the sound travel?

 

[Claymore]

so speed of sound x time = distance?

343m/s * 3s = 1029 m

correct me if I am wrong :(

 

[Kurdt]

so speed of sound x time = distance?

343m/s * 3s = 1029 m

correct me if I am wrong :(

That is correct. Now how do you think you can find the horizontal distance?

 

[Claymore]

That is correct. Now how do you think you can find the horizontal distance?

put it in Triangle :p

so 40m^2 Height x Hd^2 = 1029^2

Hd^2 = 1029^ + 1600

Hd= 32 m

correct?

 

[Kurdt]

No that's not correct. Remember that the hypotenuse of a right angled triangle is given by $a^2 = b^2 + c^2$. That means you need to subtract the vertical height squared from the hypotenuse squared and square root it. Thats a horrible sentence so I'll write it in formula.

$$b = \sqrt{a^2 - c^2}$$

Next in the problem you'll have to find the time of flight of the rock. How do you think you would find that?