Relative motion in an elevator

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Homework Help Overview

The discussion revolves around a scenario involving relative motion in an elevator. The original poster describes throwing a ball straight up and down while the elevator moves upward at a constant speed. They inquire whether an observer on the ground would see the ball reach its peak at the same time as they do and what the observed motion would look like from the ground perspective.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the implications of relative motion, questioning how the observer's perspective affects the perceived motion of the ball. Some suggest using kinematics to analyze the situation, while others discuss the differences in vertical speed as perceived by the observer and the original poster.

Discussion Status

The discussion is active, with participants providing insights and calculations related to the problem. There is acknowledgment of differing perspectives on the motion of the ball, and some participants have shared mathematical expressions to describe the time taken for the ball's motion from different frames of reference.

Contextual Notes

Participants are considering the effects of motion in a non-inertial reference frame (the elevator) and the potential implications of special relativity, although the primary focus remains on classical mechanics.

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Homework Statement


So I am in an elevator throwing a ball in an air, straight up and down, about 1 meter from my hand. The elevator is moving upwards at a rate of 2 m/s. From an observer on the ground, would they see the ball reach its peak at the same instant that I do? If not what is this difference in time.

From my perspective, I see the ball move up 1 meter, then down 1 meter. How far up and down would the observer see the ball move?

2. The attempt at a solution
My thoughts are that the observer would see the ball go higher up, as opposed to 1 m. But then they would see the ball move a shorter distance on the way down. As for the first part, I'm not sure where to begin. My guess is that the times would not be the same, but again, I'm in a stump. Thanks
 
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I think you guessed right but maybe you could prove it using kinematics?
 
Unless you are considering special relativistic effects, the time would be the same for both observer.
 
rpthomps said:
Unless you are considering special relativistic effects, the time would be the same for both observer.
That would be true if they agreed about the event, but as the OP noted, they would disagree about the vertical speed of the ball at any given instant. In particular they would disagree about whether its vertical speed is zero.
 
Ravens Fan said:

Homework Statement


So I am in an elevator throwing a ball in an air, straight up and down, about 1 meter from my hand. The elevator is moving upwards at a rate of 2 m/s. From an observer on the ground, would they see the ball reach its peak at the same instant that I do? If not what is this difference in time.

From my perspective, I see the ball move up 1 meter, then down 1 meter. How far up and down would the observer see the ball move?

2. The attempt at a solution
My thoughts are that the observer would see the ball go higher up, as opposed to 1 m. But then they would see the ball move a shorter distance on the way down. As for the first part, I'm not sure where to begin. My guess is that the times would not be the same, but again, I'm in a stump. Thanks
Why don't you do the math, and let the math answer the question for you?

Chet
 
haruspex said:
That would be true if they agreed about the event, but as the OP noted, they would disagree about the vertical speed of the ball at any given instant. In particular they would disagree about whether its vertical speed is zero.

I never considered this. It seems quite dramatic. I did some calculations and came up with the following general forms.

From inside the elevator time of object to rise and fall:

##t=\frac { 2v_o}{g }##

From outside the elevator:

##t=\frac{v_o+\sqrt { v_o^2-v_E^2 } +v_E}{g}##

Although I didn't originate the question, I learned a lot. Thanks
 

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