Calculating Initial Speed: Soccer Player Kicks Rock Off Cliff into Water

[blazin247nc]

Here's a tricky one!

Homework Statement

A soccer player kicks a rock horizontally off a 40.0m 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.

I am completely confused on where to even start. It does not seem like enough information is given in the problem but there is an answer for it in the back of the book (I just need to show my work).

I am assuming that the sound was heard 3 seconds after he kicked it (not 3 seconds after it hit the water because that would be one hell of a kick even for a soccer player). That given I would need to know the distance the rock was away from the person in order to get the time it took to relay the sound. If I had that i could find the time it took for the rock to hit the water.

What am I missing here?

 

[Doc Al]

I am assuming that the sound was heard 3 seconds after he kicked it (not 3 seconds after it hit the water because that would be one hell of a kick even for a soccer player).

Right.

That given I would need to know the distance the rock was away from the person in order to get the time it took to relay the sound. If I had that i could find the time it took for the rock to hit the water.

You have all the information needed to calculate the time it takes for the rock to hit the water. That's the first step.

 

[blazin247nc]

my original work is as follows:

40/t = 343m/s
t=0.117s <--- time the sound takes to relay

3-0.117= 2.88 <---- time for the rock to hit the water.

HOWEVER, that is only based on the distance of the y direction and doesn't incorporate how far the rock traveled in the x direction.

 

[Doc Al]

my original work is as follows:

40/t = 343m/s
t=0.117s <--- time the sound takes to relay

That would be true if the rock hit the water directly underneath the kicker. But it didn't.

3-0.117= 2.88 <---- time for the rock to hit the water.

You're attacking this backwards. First figure out the time for the rock to hit the water. (Forget about the sound for the moment.)

HOWEVER, that is only based on the distance of the y direction and doesn't incorporate how far the rock traveled in the x direction.

Right. That's why it's no good.

 

[blazin247nc]

y(final)=y(initial)+Vy(initial)*t - 1/2(9.8)(t^2)
-40= 0 + 0 -4.9t^2
8.16 = t^2
t = 2.86

Since it is projected horizontally the initial velocity in the y direction should be 0 as well as the initial y position, correct?

 

[Doc Al]

Good. Now you're cooking.

 

[blazin247nc]

Since it is projected horizontally the angle would be 0 so the initial velocity would be equal to the intial velocity in the x direction.

So from here, without the final x position I don't see how I can find the intial velocity.

 

[blazin247nc]

nm i figured it out ill post my solution in a second

 

[blazin247nc]

since the time is 2.86s then the time of the sound is .14s

343 * .14 = 48.02

48^2 - 40^2 = 704 ... 704^1/2 = 26.5 <---pythagorean theorem

x(final)=26.5

x(final)=Vx(initial)*t

Vx(initial)=26.5/2.86 = 9.3

The answer in the back of the book is 9.91 so maybe roundoff error?

 

[Doc Al]

I'd say that your answer is correct and that the book's answer is off.

 

[blazin247nc]

doc al...your a genius

thank you for your help