Tricky 2d motion question

In summary, the soccer player kicks a rock horizontally off a cliff 38.2m high into a pool of water. The sound of the splash is heard 3.13s later. To determine the initial speed of the rock, the time it takes for the rock to fall to the pool and the time it takes for the sound to reach the player's ears must first be calculated. Using this information, the horizontal displacement of the rock can be determined using the Pythagorean theorem. The initial speed of the rock can then be calculated by dividing the horizontal displacement by the fall time."
  • #1
375551
2
0
A soccer player kicks a rock horizontally off a cliff 38.2m high into a pool of water. If the player hears the sound of the splash 3.13s later, what was the initial speed given to the rock? Assume the speed of sound in air to be 343m/s

What I don't understand here is, determining the horizontal displacement to determine the initial horizontal velocity.

1) I determine the time it take the rock to fall to the pool using Viy = 0.0 m/s and using
deltax= vit +1/2at^2
2) determine the time it takes for the sound to reach the ears by subtracting 3.13-2.792 (time for rock to fall) = 0.338 s
3) now i need to determine the horizontal displacement of the rock, Dont understand how to do that.
 
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  • #2
What was the distance from the player to the splash? You have enough information to work that out, and once you do you can draw a triangle and the Pythagorean theorem will get you home from there.
 
  • #3
D=At. so I would think you would need to mass of the object to calculate the velocity on the y-axis using g so you would determine which time was taking on the y-axis and which time on the x axis. not a physics guru but now you have me trying to solve this one. is that the question word for word? nothing taken out? displacement to y-axis is D=1/2at^2. its accelerating on the y-axis at -9.8 ms^2 but still don't know the time cause we don't have the mass
 
  • #4
Try drawing a diagram of the sound returning to the kicker, treating the sound as a 'projectile.' (Edit: Assuming a straight line from source to the kicker) You have a speed and a time for the sound returning. My mental image (and gut) shows that it becomes a geometry problem at that point, seeing as you have the information about the vertical displacement already.

Edit: looks like I'm a bit late but my solution is in line with post #2. Not sure if you can do it any other way.
 
  • #5
Wouldn't kicking a rock hurt the soccer player's foot?
 
  • #6
Let the velocity be v. then the horizontal distance is D = v*(t=2.792).
Then, apply the pythagorean theorem sqrt(D^2 + h^2) = 343*0.338.
height of the cliff (known).
 
  • #7
sudu is correct.

You know the acceleration of gravity and the vertical height that the rock falls, so you know the time it took for the rock to fall.
You know that the time period from the kick to hearing the splash includes both the fall time and the time for the splash sound to return.
Knowing the speed of sound and the time from the splash to hearing it, you know the distance from the kicker to the splash.
Knowing that distance (the hypotenuse) and one of the sides (*the vertical height), you can know the other side (the horizontal distance).
Knowing the horizontal distance and the fall time, you can determine the initial speed of the rock.

*I'm assuming that the problem uses a soccer player so he can kick the rock and follow through by falling down so that his ear is at the edge of the cliff at the kick point so that the vertical component of the path of the sound back to him from the splash matches the actual height of the rock's fall to the water... and he stays down until he hears it.
Otherwise, you would need to take into account and add his height, which is not provided.
 
  • #8
Thank you all for your replys. I understand how to solve the question, I would like some more insight as to why you would assume the speed of the sound multiplied by the time for sound to travel up to be the hypotenuse. In my head I, imagine sound waves moving directly up, hence my confusion.
 
  • #9
375551 said:
Thank you all for your replys. I understand how to solve the question, I would like some more insight as to why you would assume the speed of the sound multiplied by the time for sound to travel up to be the hypotenuse. In my head I, imagine sound waves moving directly up, hence my confusion.

"Directly up" from where? Where is the sound formed? Where is it observed? What's the path from one to the other?
 
  • #10
Sound travels radially in all directions; the direction of importance here is the line between the splash point and the soccer player.
The sound of the splash originates where the rock hits the water.
The soccer player did not just push the rock over the edge of the cliff, he kicked it.
The rock did not just fall down to the water at the foot of the cliff, it traveled forward out over the water and splashed down way out in front of the cliff.
The sound has to travel both up from the water level to the cliff top, and back across the distance from the splash to the cliff - or more simply, a diagonal line from the splash point up and back to the soccer player on the cliff. That diagonal line is the distance the sound travels...
 

1. What is a "Tricky 2d motion question"?

A "Tricky 2d motion question" is a type of problem that involves motion in two dimensions, typically in the form of a graph or diagram. These types of questions are commonly found in physics and involve analyzing the motion of an object based on given information.

2. How do I approach solving a "Tricky 2d motion question"?

The best approach to solving a "Tricky 2d motion question" is to first carefully read and understand the given information and any given diagrams or graphs. Then, use the appropriate equations and principles of physics to analyze the motion and solve for the unknown variables.

3. What are some common mistakes to avoid when solving a "Tricky 2d motion question"?

Some common mistakes to avoid when solving a "Tricky 2d motion question" include using incorrect equations, not considering all forces acting on the object, and confusing horizontal and vertical components of motion. It is important to double check all calculations and make sure they align with the given information and principles of physics.

4. Can "Tricky 2d motion questions" be solved without using equations?

While equations are typically used to solve "Tricky 2d motion questions", some problems may be solved using graphical methods such as constructing vector diagrams or using the slope of a position vs. time graph. However, these methods may not be applicable to all types of "Tricky 2d motion questions".

5. How can I improve my skills in solving "Tricky 2d motion questions"?

The best way to improve your skills in solving "Tricky 2d motion questions" is to practice regularly and become familiar with the principles and equations used in these types of problems. It can also be helpful to seek guidance from a teacher or tutor and to use online resources and practice problems to improve your understanding.

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