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Physics rock off a cliff question

  • Thread starter itzela
  • Start date
  • #1
34
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Here's the problem:

A soccer player kicks a rock horizontally of a 40m high cliff into a pool of water. If the player hears the sound of the splash 3.08 s later, what was the initial speed given to the rock? Assume the speed of sound in air to be 343 m/s.

** my question is, what role does the speed of sound play in this problem?!
 

Answers and Replies

  • #2
15
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I'll take a stab at this:

Assuming air resistance is neglected or neglectable (I assume this is true unless there's something more we need to know), the time until the rock hits the water is based on the height and the acceleration due to gravity.
[tex]\frac{1}{2}gt^2 = 40[/tex] should give you the time until the rock hits.

The remainder of the time is spent with the sound waves traveling back to the listener, who is [tex]\sqrt{40^2 + x^2}[/tex] meters away, with x being the horizontal distance the rock traveled. So the distance the sound traveled is the time the sound wave traveled times the speed of sound. This gives you the horizontal distance traveled, which then gives you the initial velocity, once again assuming we can neglect the effects of air resistance on the rock.
 
  • #3
r3dxP
itzela said:
Here's the problem:

A soccer player kicks a rock horizontally of a 40m high cliff into a pool of water. If the player hears the sound of the splash 3.08 s later, what was the initial speed given to the rock? Assume the speed of sound in air to be 343 m/s.

** my question is, what role does the speed of sound play in this problem?!
there is no connection between the speed of sound in air with the question whatsoever. to solve this, u simply do.. 40=Vi(3.08s)+1/2(-9.8m/s^2)(3.08s)^2 and solve for Vi.
 
  • #4
34
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r3dxP:
i tried doing what you did before, i got 28.07m/s, but it wasn't the right answer.

StNowhere:
using your method, i got that the horizontal distance traveled (x) = 86.316, and from there i plugged in to the equation:
X = Xo + Vx*t and got Vx = 86.216/3.08 = 24.834

... is this correct? thanks a bunch!
 
  • #5
15
0
itzela said:
r3dxP:
i tried doing what you did before, i got 28.07m/s, but it wasn't the right answer.

StNowhere:
using your method, i got that the horizontal distance traveled (x) = 86.316, and from there i plugged in to the equation:
X = Xo + Vx*t and got Vx = 86.216/3.08 = 24.834

... is this correct? thanks a bunch!
I don't think we came up with the same horizontal distance. Make sure you know that there are two separate parts to the time, the time it takes for the rock to impact, and the time it takes for the sound waves of the impact to reach the kicker.
 
Last edited:
  • #6
15
0
To see how this is going, how much time did you get for the rock in the air? That should speak volumes.
 

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